High school support · California Integrated Pathway

Integrated Mathematics 2

Extra help for the Integrated Mathematics 2 class you are already in. Your teacher has covered these topics; this is where you come when the homework will not go. Every topic gives you the method as a refresher (the steps, and the specific mistake that costs the marks) then worked examples, then fifteen practice problems. Every problem has a complete worked solution, so when you get one wrong you can find the exact line where it went wrong instead of only knowing that it did.

INTEGRATED PATHWAY CA CCSS MATH GRADE 10 WORKED SOLUTIONS 20 TOPICS 360 PRACTICE PROBLEMS Integrated Mathematics 1 or Algebra 1. This supports a class you are enrolled in rather than replacing it.

Course overview

Find the topic you are stuck on

This is the year most students find the steepest. Quadratics arrive with four competing solution methods and no obvious way to choose between them, similarity and circle theorems demand proof rather than computation, and probability asks for careful counting where every other topic has asked for algebra. The topics below follow the order most California Integrated II courses use, so you can go straight to whatever was covered in class today.

  • U1Unit 1: Polynomials and Factoring3 topics
  • U2Unit 2: Quadratic Functions and Equations4 topics
  • U3Unit 3: Similarity and Right Triangles4 topics
  • U4Unit 4: Circles3 topics
  • U5Unit 5: Measurement and Modeling3 topics
  • U6Unit 6: Probability3 topics

All six units are open, 20 topics in all. Every topic opens with the method, worked examples, and fifteen practice problems. Every problem has a full worked solution, so you can find the step where yours went wrong. Each unit closes with a ten-problem mixed review.

Free preview: open any 5 topics without an account. The counter on the left keeps track.

Topic 1.1 · Unit 1 · CA CCSS N-RN.1–2, 8-EE.1

Exponent rules and scientific notation

Every exponent rule comes from one fact: an exponent counts how many copies are multiplied. If you forget a rule, expand a small case and read the answer off; that takes ten seconds and never misleads you.

The method
  1. Multiplying, same base: add the exponents. \( x^3 \cdot x^4 = x^7 \), because three copies times four copies is seven copies.
  2. Dividing, same base: subtract. \( \frac{x^7}{x^4} = x^3 \).
  3. Power of a power: multiply. \( (x^3)^4 = x^{12} \), four groups of three copies.
  4. Power of a product: the exponent reaches everything inside. \( (2x)^3 = 8x^3 \), not \( 2x^3 \). The 2 is cubed as well.
  5. Zero and negative: \( x^0 = 1 \) for any \( x \neq 0 \), and \( x^{-n} = \frac{1}{x^n} \). A negative exponent means reciprocal, never a negative answer.
  6. Scientific notation: \( a \times 10^n \) with \( 1 \le a < 10 \). Multiply by adding exponents and multiplying the fronts; if the front leaves that range, fix it by shifting one place.

Where marks are lost: \( (-3)^2 = 9 \) but \( -3^2 = -9 \), without brackets the exponent binds tighter than the minus. And \( 2^{-3} = \frac{1}{8} \), a positive number; students frequently write \( -8 \).

Worked examples

Example 1: combining rules. Simplify \( \frac{(2x^3)^2 \cdot x^4}{4x^5} \).

Handle the bracket first: \( (2x^3)^2 = 4x^6 \), the 2 is squared too. Numerator: \( 4x^6 \cdot x^4 = 4x^{10} \). Now divide: \( \frac{4x^{10}}{4x^5} = x^5 \).

Example 2: a negative exponent. Write \( \frac{3x^{-2}}{y^{-4}} \) with positive exponents only.

A factor with a negative exponent moves across the fraction bar and the sign flips. The \( x^{-2} \) goes down, the \( y^{-4} \) comes up: \( \frac{3y^4}{x^2} \). Note the 3 does not move; it has no negative exponent.

Example 3: scientific notation. Compute \( (4 \times 10^5)(3 \times 10^{-8}) \).

Multiply the fronts and add the exponents: \( 12 \times 10^{-3} \). But 12 is not between 1 and 10, so shift one place: \( 1.2 \times 10^{-2} \). Making the front smaller means making the exponent bigger, so the value is unchanged.

Practice · 15 problems

1–6 apply one rule, 7–12 combine them, 13–15 use scientific notation and diagnosis.

  1. Simplify \( x^5 \cdot x^3 \).
    Show the full solution

    Same base, so add: \( x^8 \)

  2. Simplify \( \frac{y^9}{y^4} \).
    Show the full solution

    Subtract: \( y^5 \)

  3. Simplify \( (a^4)^3 \).
    Show the full solution

    Multiply: \( a^{12} \)

  4. Simplify \( (3x)^2 \).
    Show the full solution

    The exponent reaches the 3 as well: \( 9x^2 \)

  5. Evaluate \( 7^0 \).
    Show the full solution

    Any non-zero base to the zero power. 1

  6. Write \( 4^{-2} \) as a fraction.
    Show the full solution

    \( \frac{1}{4^2} = \frac{1}{16} \). Positive, not negative. \( \frac{1}{16} \)

  7. Simplify \( (2x^3)^4 \).
    Show the full solution

    \( 2^4 = 16 \) and \( (x^3)^4 = x^{12} \). \( 16x^{12} \)

  8. Simplify \( \frac{12x^7}{3x^2} \).
    Show the full solution

    Divide the numbers, subtract the exponents. \( 4x^5 \)

  9. Simplify \( \frac{(x^2)^5}{x^3} \).
    Show the full solution

    \( x^{10} \div x^3 \). \( x^7 \)

  10. Write \( \frac{5x^{-3}}{y^{-2}} \) with positive exponents.
    Show the full solution

    \( x^{-3} \) moves down, \( y^{-2} \) moves up; the 5 stays. \( \frac{5y^2}{x^3} \)

  11. Simplify \( (4x^2y)(3xy^3) \).
    Show the full solution

    Numbers: \( 12 \). \( x \): \( 2 + 1 = 3 \). \( y \): \( 1 + 3 = 4 \). \( 12x^3y^4 \)

  12. Simplify \( \left( \frac{x^3}{2} \right)^2 \).
    Show the full solution

    Square top and bottom: \( \frac{x^6}{4} \). \( \frac{x^6}{4} \)

  13. Compute \( (5 \times 10^4)(6 \times 10^3) \) in scientific notation.
    Show the full solution

    \( 30 \times 10^7 \); 30 is out of range, so shift: \( 3 \times 10^8 \). \( 3 \times 10^8 \)

  14. A student writes \( -4^2 = 16 \). Find the error.
    Show the full solution

    Without brackets the exponent applies only to the 4, and the minus is applied after: \( -(4^2) = -16 \). It would be 16 only if written \( (-4)^2 \). \( -16 \)

  15. A student writes \( (3x)^2 = 3x^2 \). Find the error.
    Show the full solution

    The exponent applies to everything inside the bracket, the 3 included: \( 3^2 x^2 = 9x^2 \). Check with \( x = 1 \): \( (3)^2 = 9 \), while their version gives 3. \( 9x^2 \)

Topic 1.2 · Unit 1 · CA CCSS A-APR.1

Adding, subtracting and multiplying polynomials

Adding and subtracting polynomials is combining like terms with one trap, the subtraction sign. Multiplying is the distributive property applied until nothing is left in brackets.

The method
  1. Adding: combine like terms. Only identical variable parts combine, so \( 3x^2 \) and \( 3x \) stay separate.
  2. Subtracting: distribute the minus to every term in the second polynomial, then add. This single step accounts for most errors in the topic.
  3. Multiplying a monomial by a polynomial: distribute to each term, multiplying numbers and adding exponents.
  4. Multiplying two binomials: every term in the first multiplies every term in the second, four products for two binomials. FOIL is a name for this order, not a separate rule, and it only works for binomial times binomial.
  5. Two patterns worth recognizing instantly: \( (a+b)(a-b) = a^2 - b^2 \), the middle terms cancel. And \( (a+b)^2 = a^2 + 2ab + b^2 \), not \( a^2 + b^2 \).

Where marks are lost: \( (x+3)^2 \) is not \( x^2 + 9 \). Squaring a binomial means multiplying it by itself, which produces a middle term: \( x^2 + 6x + 9 \). Test with \( x = 1 \): \( 4^2 = 16 \), and \( 1 + 6 + 9 = 16 \) ✓, while \( 1 + 9 = 10 \) ✗.

Worked examples

Example 1: subtraction. Simplify \( (5x^2 - 3x + 4) - (2x^2 + x - 7) \).

Distribute the minus across all three terms of the second: \( -2x^2 - x + 7 \). Now combine: \( 5x^2 - 2x^2 = 3x^2 \); \( -3x - x = -4x \); \( 4 + 7 = 11 \).

Result \( 3x^2 - 4x + 11 \). The \( +7 \) is where marks go; it was \( -7 \) inside the bracket.

Example 2: two binomials. Expand \( (2x + 3)(x - 5) \).

Four products: \( 2x \cdot x = 2x^2 \); \( 2x \cdot -5 = -10x \); \( 3 \cdot x = 3x \); \( 3 \cdot -5 = -15 \).

Combining the middle terms: \( 2x^2 - 7x - 15 \).

Example 3: the difference of squares pattern. Expand \( (x + 6)(x - 6) \).

The middle products are \( -6x \) and \( +6x \), which cancel, leaving \( x^2 - 36 \). Recognizing this pattern going forwards makes factoring it backwards in the next topic much faster.

Practice · 15 problems

1–5 add and subtract, 6–12 multiply, 13–15 use the patterns and diagnosis.

  1. Simplify \( (3x + 5) + (2x - 8) \).
    Show the full solution

    \( 5x - 3 \). \( 5x - 3 \)

  2. Simplify \( (4x^2 + x) + (x^2 - 3x) \).
    Show the full solution

    \( 5x^2 - 2x \). \( 5x^2 - 2x \)

  3. Simplify \( (7x - 2) - (3x + 6) \).
    Show the full solution

    Distribute the minus: \( -3x - 6 \). Then \( 4x - 8 \). \( 4x - 8 \)

  4. Simplify \( (x^2 - 4x + 1) - (2x^2 - x - 5) \).
    Show the full solution

    Second becomes \( -2x^2 + x + 5 \). Combining: \( -x^2 - 3x + 6 \). \( -x^2 - 3x + 6 \)

  5. Simplify \( (6x^2 + 2x - 9) - (6x^2 - 2x + 9) \).
    Show the full solution

    Second becomes \( -6x^2 + 2x - 9 \). The \( x^2 \) terms cancel: \( 4x - 18 \). \( 4x - 18 \)

  6. Expand \( 3x(x + 4) \).
    Show the full solution

    \( 3x^2 + 12x \)

  7. Expand \( -2x(3x - 5) \).
    Show the full solution

    \( -6x^2 + 10x \). The second sign flips. \( -6x^2 + 10x \)

  8. Expand \( (x + 2)(x + 7) \).
    Show the full solution

    \( x^2 + 7x + 2x + 14 \). \( x^2 + 9x + 14 \)

  9. Expand \( (x - 3)(x + 8) \).
    Show the full solution

    \( x^2 + 8x - 3x - 24 \). \( x^2 + 5x - 24 \)

  10. Expand \( (2x + 1)(3x - 4) \).
    Show the full solution

    \( 6x^2 - 8x + 3x - 4 \). \( 6x^2 - 5x - 4 \)

  11. Expand \( (x - 5)(x + 5) \).
    Show the full solution

    Middle terms cancel. \( x^2 - 25 \)

  12. Expand \( (x + 4)^2 \).
    Show the full solution

    \( (x+4)(x+4) = x^2 + 4x + 4x + 16 \). \( x^2 + 8x + 16 \)

  13. Expand \( (3x - 2)^2 \).
    Show the full solution

    \( 9x^2 - 6x - 6x + 4 \). \( 9x^2 - 12x + 4 \)

  14. A student writes \( (x + 5)^2 = x^2 + 25 \). Find the error.
    Show the full solution

    They squared each term instead of multiplying the binomial by itself, losing the middle term: \( x^2 + 10x + 25 \). Test \( x = 1 \): \( 6^2 = 36 \), and \( 1 + 10 + 25 = 36 \) ✓, while their answer gives 26. \( x^2 + 10x + 25 \)

  15. A student simplifies \( (5x - 1) - (2x - 4) \) to \( 3x - 5 \). Find the error.
    Show the full solution

    They subtracted the \( 2x \) correctly but kept the \( -4 \) negative. Distributing the minus gives \( -2x + 4 \), so the answer is \( 3x + 3 \). \( 3x + 3 \)

Topic 1.3 · Unit 1 · CA CCSS A-SSE.2–3

Factoring polynomials

Factoring is expanding run backwards, and there is a fixed order to try things in. Working through that order rather than guessing is what makes this reliable, and it is the skill the whole quadratics unit depends on.

The method

Always in this order:

  1. Greatest common factor first, every time. Pull out what every term shares. \( 6x^2 + 9x = 3x(2x + 3) \). Skipping this makes everything after it harder and sometimes impossible.
  2. Count the terms. Two terms → look for a difference of squares: \( a^2 - b^2 = (a+b)(a-b) \). Note \( a^2 + b^2 \) does not factor.
  3. Three terms, leading coefficient 1: for \( x^2 + bx + c \), find two numbers that multiply to \( c \) and add to \( b \). Those two numbers go straight into \( (x + \_)(x + \_) \).
  4. Three terms, leading coefficient not 1: for \( ax^2 + bx + c \), find two numbers multiplying to \( a \times c \) and adding to \( b \), split the middle term into those two, then factor in pairs.
  5. Check by expanding. Factoring is the one topic where checking is instant and catches nearly every error.

Signs, which is where most marks go: if \( c \) is positive, both numbers share the sign of \( b \). If \( c \) is negative, the numbers have opposite signs and the larger one carries the sign of \( b \).

Worked examples

Example 1: GCF then a pattern. Factor \( 2x^3 - 18x \).

GCF is \( 2x \): \( 2x(x^2 - 9) \). The bracket is now two terms and both are perfect squares, so it is a difference of squares: \( 2x(x+3)(x-3) \).

Pulling the GCF first is what revealed the pattern, without it, \( 2x^3 - 18x \) looks like nothing.

Example 2: trinomial with leading coefficient 1. Factor \( x^2 - 5x - 24 \).

Need two numbers multiplying to \( -24 \) and adding to \( -5 \). Since the product is negative the signs differ; since the sum is negative the larger number is negative. Pairs for 24: 1 and 24, 2 and 12, 3 and 8, 4 and 6. The pair 3 and 8 differs by 5, so take \( +3 \) and \( -8 \).

\( (x + 3)(x - 8) \). Check: \( -8x + 3x = -5x \) ✓ and \( 3 \times -8 = -24 \) ✓.

Example 3: leading coefficient not 1. Factor \( 3x^2 + 11x + 6 \).

\( a \times c = 18 \), and we need two numbers multiplying to 18 and adding to 11: that is 9 and 2. Split the middle term:

\( 3x^2 + 9x + 2x + 6 \). Factor in pairs: \( 3x(x + 3) + 2(x + 3) \). Both share \( (x+3) \), so it factors as \( (x + 3)(3x + 2) \).

Check by expanding: \( 3x^2 + 2x + 9x + 6 = 3x^2 + 11x + 6 \) ✓.

Practice · 15 problems

1–5 GCF and patterns, 6–11 trinomials, 12–15 harder cases and diagnosis.

  1. Factor \( 4x + 12 \).
    Show the full solution

    GCF 4. \( 4(x + 3) \)

  2. Factor \( 6x^2 - 15x \).
    Show the full solution

    GCF \( 3x \). \( 3x(2x - 5) \)

  3. Factor \( x^2 - 49 \).
    Show the full solution

    Difference of squares. \( (x+7)(x-7) \)

  4. Factor \( 9x^2 - 25 \).
    Show the full solution

    Both perfect squares: \( (3x)^2 - 5^2 \). \( (3x+5)(3x-5) \)

  5. Can \( x^2 + 16 \) be factored over the real numbers?
    Show the full solution

    No. A sum of squares does not factor; only the difference does. Does not factor

  6. Factor \( x^2 + 7x + 12 \).
    Show the full solution

    Two numbers multiplying to 12, adding to 7: 3 and 4. \( (x+3)(x+4) \)

  7. Factor \( x^2 - 9x + 20 \).
    Show the full solution

    Product positive and sum negative, so both negative: \( -4 \) and \( -5 \). \( (x-4)(x-5) \)

  8. Factor \( x^2 + 2x - 15 \).
    Show the full solution

    Product negative so signs differ; \( +5 \) and \( -3 \) add to \( +2 \). \( (x+5)(x-3) \)

  9. Factor \( x^2 - 3x - 28 \).
    Show the full solution

    Signs differ, larger negative: \( -7 \) and \( +4 \). \( (x-7)(x+4) \)

  10. Factor \( x^2 + 10x + 25 \).
    Show the full solution

    5 and 5. \( (x+5)^2 \)

  11. Factor \( 2x^2 + 14x + 24 \).
    Show the full solution

    GCF 2 first: \( 2(x^2 + 7x + 12) \), then the trinomial. \( 2(x+3)(x+4) \)

  12. Factor \( 2x^2 + 7x + 3 \).
    Show the full solution

    \( ac = 6 \); need two numbers multiplying to 6, adding to 7: 6 and 1. Split: \( 2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3) \). \( (x+3)(2x+1) \)

  13. Factor \( 3x^2 - 10x + 8 \).
    Show the full solution

    \( ac = 24 \); two numbers multiplying to 24 and adding to \( -10 \): \( -6 \) and \( -4 \). Split: \( 3x^2 - 6x - 4x + 8 = 3x(x-2) - 4(x-2) \). \( (x-2)(3x-4) \)

  14. A student factors \( x^2 - 16 \) as \( (x-4)^2 \). Find the error.
    Show the full solution

    \( (x-4)^2 \) expands to \( x^2 - 8x + 16 \), which has a middle term and the wrong sign on 16. A difference of squares factors into a sum times a difference: \( (x+4)(x-4) \). \( (x+4)(x-4) \)

  15. A student factors \( 2x^2 + 10x \) as \( (2x)(x + 5) \) and stops. Is that fully factored?
    Show the full solution

    It is correct but the GCF was not fully taken; it is conventionally written \( 2x(x+5) \), which is the same thing. The expression is fully factored either way, since \( x + 5 \) cannot factor further. Correct; write it as \( 2x(x+5) \)

Unit 1 mixed review · 10 problems · all topics

Unit 1 mixed review: Polynomials and Factoring

These are shuffled across all three topics and do not tell you which method they want, which is what makes them closer to a real test than a single topic's practice set.

  1. Simplify \( (3x^2y)(4x^5y^3) \).
    Show the full solution

    Multiply coefficients, add exponents on each base: \( 12x^7y^4 \). \( 12x^7y^4 \)

  2. Simplify \( \frac{20a^6b^2}{5a^2b^5} \).
    Show the full solution

    \( \frac{20}{5} = 4 \), \( a^{6-2} = a^4 \), \( b^{2-5} = b^{-3} \). \( \frac{4a^4}{b^3} \)

  3. Evaluate \( \left(\frac{2}{3}\right)^{-2} \).
    Show the full solution

    A negative exponent flips the fraction: \( \left(\frac{3}{2}\right)^2 = \frac{9}{4} \). \( \frac{9}{4} \)

  4. Expand \( (2x - 5)(3x + 4) \).
    Show the full solution

    \( 6x^2 + 8x - 15x - 20 = 6x^2 - 7x - 20 \). \( 6x^2 - 7x - 20 \)

  5. Expand \( (4x - 3)^2 \).
    Show the full solution

    \( 16x^2 - 24x + 9 \). The middle term is \( 2(4x)(-3) \), omitting it is the standard error. \( 16x^2 - 24x + 9 \)

  6. Factor \( x^2 - 11x + 24 \).
    Show the full solution

    Two numbers multiplying to 24 and adding to \( -11 \): \( -3 \) and \( -8 \). \( (x-3)(x-8) \)

  7. Factor \( 9x^2 - 49 \).
    Show the full solution

    Difference of squares. \( (3x-7)(3x+7) \)

  8. Factor \( 6x^2 + 11x - 10 \).
    Show the full solution

    \( ac = -60 \); a pair giving \( +11 \) is \( 15 \) and \( -4 \). \( 6x^2 + 15x - 4x - 10 = 3x(2x+5) - 2(2x+5) \). \( (3x-2)(2x+5) \)

  9. Factor completely: \( 4x^3 - 36x \).
    Show the full solution

    Common factor first: \( 4x(x^2 - 9) \), then difference of squares. Stopping at \( 4x(x^2-9) \) is the usual lost mark. \( 4x(x-3)(x+3) \)

  10. A student writes \( (x+3)^2 = x^2 + 9 \). Explain the error and correct it.
    Show the full solution

    Squaring is not distributive over addition. \( (x+3)^2 = (x+3)(x+3) \), which gives \( x^2 + 3x + 3x + 9 \). The cross terms are what was lost. \( x^2 + 6x + 9 \)

Topic 2.1 · Unit 2 · CA CCSS A-REI.4b

Solving quadratics by factoring and square roots

A quadratic usually has two solutions, and the fastest route depends on its shape. If there is no middle term, take square roots. If it factors, factor. Both methods rest on one idea worth stating properly.

The method
  1. The zero product property: if \( A \cdot B = 0 \) then \( A = 0 \) or \( B = 0 \). This is why factoring solves equations, and it only works against zero. From \( (x-2)(x-3) = 6 \) you may not conclude \( x - 2 = 6 \); expand and rearrange to zero first.
  2. To solve by factoring: move everything to one side so the other is 0, factor, set each factor to zero, solve each.
  3. To solve by square roots when there is no \( x \) term: isolate the squared quantity, then take roots, and write ±. Forgetting it loses half the answer.
  4. Check the count. Two distinct solutions, one repeated solution (a perfect square), or none in the reals (when a square equals a negative).
  5. In context, discard impossible answers. Negative lengths and negative times are not solutions to the real problem, and saying so explicitly earns the mark.

Where marks are lost: dividing both sides by \( x \). From \( x^2 = 5x \), dividing by \( x \) gives \( x = 5 \) and silently destroys the solution \( x = 0 \). Move everything across instead: \( x^2 - 5x = 0 \), so \( x(x-5) = 0 \) and \( x = 0 \) or \( x = 5 \).

Worked examples

Example 1: rearranging before factoring. Solve \( x^2 + 3x = 10 \).

Move everything to one side: \( x^2 + 3x - 10 = 0 \). Two numbers multiplying to \( -10 \) and adding to 3 are 5 and \( -2 \), so \( (x+5)(x-2) = 0 \).

Therefore \( x = -5 \) or \( x = 2 \). Check \( x = 2 \): \( 4 + 6 = 10 \) ✓.

Example 2: square roots with ±. Solve \( 2(x - 3)^2 = 50 \).

Isolate the square: \( (x-3)^2 = 25 \). Take roots, keeping both signs: \( x - 3 = \pm 5 \).

So \( x = 8 \) or \( x = -2 \). Writing only \( x - 3 = 5 \) would lose the second solution entirely.

Example 3: context. A rectangular garden is 3 m longer than it is wide and has area 40 m². Find its width.

Let the width be \( w \); the length is \( w + 3 \). Then \( w(w+3) = 40 \), so \( w^2 + 3w - 40 = 0 \), giving \( (w+8)(w-5) = 0 \) and \( w = -8 \) or \( w = 5 \).

A width cannot be negative, so \( w = 5 \) m. Both algebraic solutions are correct; only one answers the question, and saying why is part of the answer.

Practice · 15 problems

1–5 already factored or simple, 6–11 need rearranging, 12–15 are context and diagnosis.

  1. Solve \( (x - 4)(x + 7) = 0 \).
    Show the full solution

    Each factor to zero. \( x = 4 \) or \( x = -7 \)

  2. Solve \( x^2 = 36 \).
    Show the full solution

    Square roots, both signs. \( x = \pm 6 \)

  3. Solve \( x^2 - 81 = 0 \).
    Show the full solution

    Either as a difference of squares or by roots. \( x = \pm 9 \)

  4. Solve \( 3x^2 = 27 \).
    Show the full solution

    \( x^2 = 9 \). \( x = \pm 3 \)

  5. Solve \( (x + 2)^2 = 49 \).
    Show the full solution

    \( x + 2 = \pm 7 \). \( x = 5 \) or \( x = -9 \)

  6. Solve \( x^2 + 5x + 6 = 0 \).
    Show the full solution

    \( (x+2)(x+3) = 0 \). \( x = -2 \) or \( x = -3 \)

  7. Solve \( x^2 - 7x + 10 = 0 \).
    Show the full solution

    \( (x-2)(x-5) = 0 \). \( x = 2 \) or \( x = 5 \)

  8. Solve \( x^2 + 2x = 24 \).
    Show the full solution

    \( x^2 + 2x - 24 = 0 \), so \( (x+6)(x-4) = 0 \). \( x = -6 \) or \( x = 4 \)

  9. Solve \( x^2 = 7x \).
    Show the full solution

    Do not divide by \( x \). Move across: \( x^2 - 7x = 0 \), so \( x(x-7) = 0 \). \( x = 0 \) or \( x = 7 \)

  10. Solve \( 2x^2 + 7x + 3 = 0 \).
    Show the full solution

    Factors as \( (x+3)(2x+1) = 0 \). \( x = -3 \) or \( x = -\frac{1}{2} \)

  11. Solve \( x^2 - 6x + 9 = 0 \) and say how many distinct solutions there are.
    Show the full solution

    \( (x-3)^2 = 0 \), so \( x = 3 \) twice, a repeated root. One distinct solution, \( x = 3 \)

  12. Solve \( x^2 + 4 = 0 \) over the real numbers.
    Show the full solution

    \( x^2 = -4 \), and no real number squares to a negative. No real solutions

  13. A ball's height is \( h = -5t^2 + 20t \) meters after \( t \) seconds. When does it hit the ground?
    Show the full solution

    Ground means \( h = 0 \): \( -5t(t - 4) = 0 \), so \( t = 0 \) or \( t = 4 \). \( t = 0 \) is the throw itself, so it lands at \( t = 4 \) seconds.

  14. A student solves \( (x-2)(x-3) = 6 \) by writing \( x - 2 = 6 \) and \( x - 3 = 6 \). Find the error.
    Show the full solution

    The zero product property applies only against zero. Expand and rearrange: \( x^2 - 5x + 6 = 6 \), so \( x^2 - 5x = 0 \) and \( x(x-5) = 0 \). \( x = 0 \) or \( x = 5 \)

  15. A student solves \( x^2 = 9x \) by dividing both sides by \( x \) and answers \( x = 9 \). What was lost?
    Show the full solution

    Dividing by \( x \) assumes \( x \neq 0 \), discarding the solution \( x = 0 \). Factoring keeps both: \( x(x-9) = 0 \). \( x = 0 \) or \( x = 9 \)

Topic 2.2 · Unit 2 · CA CCSS A-REI.4a–b

Completing the square and the quadratic formula

Most quadratics do not factor with whole numbers. The quadratic formula solves every one of them, and it comes from completing the square, which is also how you find a vertex, so it is worth understanding rather than only memorizing.

The method
  1. The formula: for \( ax^2 + bx + c = 0 \), \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Write the equation in standard form first and read off \( a \), \( b \), \( c \) with their signs.
  2. The discriminant \( b^2 - 4ac \) tells you the answer before you finish: positive → two real solutions; zero → one repeated; negative → no real solutions.
  3. Completing the square: with \( a = 1 \), take half of \( b \), square it, and add it to both sides. That makes the left side \( (x + \frac{b}{2})^2 \), and then square roots finish it.
  4. Simplify surds by pulling out perfect squares: \( \sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2} \). Leave answers exact unless a decimal is asked for.
  5. Choosing a method: try factoring first; it is fastest when it works. Use the formula when it does not. Use completing the square when you need the vertex.

Where marks are lost: the sign of \( b \) and the whole denominator. In \( x^2 - 6x + 5 \), \( b = -6 \), so \( -b = +6 \). And the \( 2a \) divides everything above it, not just the square root.

Worked examples

Example 1: the formula with a negative \( b \). Solve \( x^2 - 6x + 5 = 0 \).

\( a = 1 \), \( b = -6 \), \( c = 5 \). Discriminant: \( (-6)^2 - 4(1)(5) = 36 - 20 = 16 \), positive, so two real solutions.

\( x = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2} \), giving \( x = 5 \) or \( x = 1 \). This one factored anyway (\( (x-5)(x-1) \)) which is a useful check that the formula was applied correctly.

Example 2: an irrational answer. Solve \( 2x^2 + 3x - 4 = 0 \).

\( a = 2 \), \( b = 3 \), \( c = -4 \). Discriminant: \( 9 - 4(2)(-4) = 9 + 32 = 41 \). Note \( -4ac \) became \( +32 \) because \( c \) is negative.

\( x = \frac{-3 \pm \sqrt{41}}{4} \). Since 41 has no square factors this is fully simplified, and it does not factor, which is exactly when the formula earns its keep.

Example 3: completing the square. Solve \( x^2 + 8x - 3 = 0 \) by completing the square.

Move the constant: \( x^2 + 8x = 3 \). Half of 8 is 4; \( 4^2 = 16 \). Add 16 to both sides:

\( x^2 + 8x + 16 = 19 \), so \( (x + 4)^2 = 19 \) and \( x + 4 = \pm\sqrt{19} \).

\( x = -4 \pm \sqrt{19} \). The form \( (x+4)^2 = 19 \) also tells you the vertex is at \( x = -4 \), which the formula alone would not show as directly.

Practice · 15 problems

1–5 discriminants, 6–11 the formula, 12–15 completing the square and diagnosis.

  1. Find the discriminant of \( x^2 + 5x + 6 = 0 \) and say how many real solutions there are.
    Show the full solution

    \( 25 - 24 = 1 \), positive. Two real solutions

  2. Find the discriminant of \( x^2 - 4x + 4 = 0 \).
    Show the full solution

    \( 16 - 16 = 0 \). One repeated solution

  3. Find the discriminant of \( x^2 + x + 5 = 0 \).
    Show the full solution

    \( 1 - 20 = -19 \), negative. No real solutions

  4. For \( 3x^2 - 2x - 8 = 0 \), state \( a \), \( b \) and \( c \).
    Show the full solution

    Signs belong to the coefficients. \( a = 3, b = -2, c = -8 \)

  5. Simplify \( \sqrt{72} \).
    Show the full solution

    \( \sqrt{36 \cdot 2} = 6\sqrt{2} \). \( 6\sqrt{2} \)

  6. Solve \( x^2 + 3x - 10 = 0 \) using the formula.
    Show the full solution

    Discriminant \( 9 + 40 = 49 \). \( x = \frac{-3 \pm 7}{2} \). \( x = 2 \) or \( x = -5 \)

  7. Solve \( x^2 - 2x - 5 = 0 \).
    Show the full solution

    Discriminant \( 4 + 20 = 24 \). \( x = \frac{2 \pm \sqrt{24}}{2} = \frac{2 \pm 2\sqrt{6}}{2} = 1 \pm \sqrt{6} \). \( x = 1 \pm \sqrt{6} \)

  8. Solve \( 2x^2 - 5x + 1 = 0 \).
    Show the full solution

    Discriminant \( 25 - 8 = 17 \). \( x = \frac{5 \pm \sqrt{17}}{4} \)

  9. Solve \( x^2 + 4x + 1 = 0 \).
    Show the full solution

    Discriminant \( 16 - 4 = 12 \). \( x = \frac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3} \). \( x = -2 \pm \sqrt{3} \)

  10. Solve \( 3x^2 + 2x - 1 = 0 \).
    Show the full solution

    Discriminant \( 4 + 12 = 16 \). \( x = \frac{-2 \pm 4}{6} \), giving \( \frac{2}{6} = \frac{1}{3} \) or \( -1 \). \( x = \frac{1}{3} \) or \( x = -1 \)

  11. Solve \( x^2 - 6x + 11 = 0 \).
    Show the full solution

    Discriminant \( 36 - 44 = -8 \), negative. No real solutions

  12. Complete the square: \( x^2 + 6x = 7 \).
    Show the full solution

    Half of 6 is 3; \( 3^2 = 9 \). Add to both sides: \( (x+3)^2 = 16 \), so \( x + 3 = \pm 4 \). \( x = 1 \) or \( x = -7 \)

  13. Complete the square: \( x^2 - 10x + 2 = 0 \).
    Show the full solution

    \( x^2 - 10x = -2 \); half of \( -10 \) is \( -5 \), squared is 25. \( (x-5)^2 = 23 \), so \( x = 5 \pm \sqrt{23} \). \( x = 5 \pm \sqrt{23} \)

  14. A student uses the formula on \( x^2 - 6x + 5 = 0 \) and writes \( x = \frac{-6 \pm \sqrt{16}}{2} \). Find the error.
    Show the full solution

    Here \( b = -6 \), so \( -b = +6 \), not \( -6 \). Correct: \( x = \frac{6 \pm 4}{2} \), giving 5 and 1. Their version gives \( -1 \) and \( -5 \), which fail when substituted back. \( x = 5 \) or \( x = 1 \)

  15. A student writes \( x = -3 \pm \frac{\sqrt{20}}{2} \) for \( x = \frac{-6 \pm \sqrt{20}}{2} \). Is that right?
    Show the full solution

    Yes, dividing both terms of the numerator by 2 gives \( -3 \pm \frac{\sqrt{20}}{2} \), which is valid. It simplifies further: \( \sqrt{20} = 2\sqrt{5} \), so \( x = -3 \pm \sqrt{5} \). The error to avoid is dividing only one term. Correct; simplifies to \( -3 \pm \sqrt{5} \)

Topic 2.3 · Unit 2 · CA CCSS F-IF.7a, F-IF.8a, F-BF.3

Graphing quadratic functions

Every quadratic graph is a parabola, and four features describe it completely: which way it opens, where its vertex sits, where it crosses the axes, and how wide it is. Each form of the equation hands you a different one of those for free.

The method
  1. Direction: \( a > 0 \) opens upward with a minimum; \( a < 0 \) opens downward with a maximum. Read the sign before anything else.
  2. Vertex form \( y = a(x-h)^2 + k \) gives the vertex \( (h, k) \) directly, with the same backwards horizontal shift as every other function.
  3. Standard form \( y = ax^2 + bx + c \): the axis of symmetry is \( x = -\frac{b}{2a} \); substitute that back in to get the \( y \)-coordinate of the vertex. The \( y \)-intercept is simply \( c \).
  4. Factored form \( y = a(x - r_1)(x - r_2) \) gives the \( x \)-intercepts directly, and the vertex sits exactly halfway between them.
  5. Width: \( |a| > 1 \) is narrower than \( y = x^2 \); \( |a| < 1 \) is wider. Parabolas are symmetric about the axis, so plotting points on one side gives the other for free.

Where marks are lost: the axis of symmetry formula is \( -\frac{b}{2a} \), and the minus is part of it. For \( y = x^2 - 4x + 1 \), \( b = -4 \), so the axis is \( -\frac{-4}{2} = 2 \), not \( -2 \).

Worked examples

Example 1: from standard form. Describe the graph of \( y = x^2 - 6x + 5 \).

\( a = 1 > 0 \), so it opens upward. Axis of symmetry: \( x = -\frac{-6}{2(1)} = 3 \). Vertex \( y \)-value: \( 9 - 18 + 5 = -4 \), so the vertex is \( (3, -4) \), a minimum.

\( y \)-intercept is \( c = 5 \). For \( x \)-intercepts, factor: \( (x-5)(x-1) = 0 \), giving \( x = 1 \) and \( x = 5 \). Note the vertex at \( x = 3 \) is exactly halfway between them, as it must be.

Example 2: from vertex form. Describe \( y = -2(x + 1)^2 + 8 \).

\( x + 1 = x - (-1) \), so \( h = -1 \) and the vertex is \( (-1, 8) \). Since \( a = -2 \) it opens downward (so 8 is a maximum) and it is twice as steep as \( y = x^2 \).

Example 3: from factored form. Find the vertex of \( y = (x - 2)(x + 6) \).

The \( x \)-intercepts are 2 and \( -6 \). The vertex lies halfway between: \( \frac{2 + (-6)}{2} = -2 \). Substituting: \( y = (-4)(4) = -16 \), so the vertex is \( (-2, -16) \).

This is faster than expanding to standard form and using \( -\frac{b}{2a} \), and it is available whenever the quadratic factors.

Practice · 15 problems

1–5 read features off, 6–11 compute them, 12–15 combine and diagnose.

  1. Does \( y = 3x^2 - 5 \) open upward or downward?
    Show the full solution

    \( a = 3 > 0 \). Upward

  2. Does \( y = -x^2 + 4x \) open upward or downward?
    Show the full solution

    \( a = -1 < 0 \). Downward, with a maximum

  3. State the vertex of \( y = (x - 4)^2 + 3 \).
    Show the full solution

    \( (4, 3) \)

  4. State the vertex of \( y = 2(x + 5)^2 - 1 \).
    Show the full solution

    \( x + 5 = x - (-5) \). \( (-5, -1) \)

  5. State the \( y \)-intercept of \( y = x^2 + 7x - 12 \).
    Show the full solution

    The constant term. \( (0, -12) \)

  6. Find the axis of symmetry of \( y = x^2 - 8x + 2 \).
    Show the full solution

    \( x = -\frac{-8}{2} = 4 \). \( x = 4 \)

  7. Find the vertex of \( y = x^2 - 8x + 2 \).
    Show the full solution

    At \( x = 4 \): \( 16 - 32 + 2 = -14 \). \( (4, -14) \)

  8. Find the axis of symmetry of \( y = 2x^2 + 12x - 1 \).
    Show the full solution

    \( x = -\frac{12}{4} = -3 \). \( x = -3 \)

  9. Find the \( x \)-intercepts of \( y = x^2 - 9 \).
    Show the full solution

    \( (x+3)(x-3) = 0 \). \( x = 3 \) and \( x = -3 \)

  10. Find the \( x \)-intercepts of \( y = (x + 1)(x - 7) \).
    Show the full solution

    \( x = -1 \) and \( x = 7 \)

  11. Find the vertex of \( y = (x + 1)(x - 7) \).
    Show the full solution

    Halfway between \( -1 \) and 7 is \( x = 3 \). Then \( y = (4)(-4) = -16 \). \( (3, -16) \)

  12. Is \( y = \frac{1}{4}x^2 \) wider or narrower than \( y = x^2 \)?
    Show the full solution

    \( |a| < 1 \), so it rises more slowly. Wider

  13. Describe \( y = -(x - 2)^2 + 9 \) fully.
    Show the full solution

    Opens downward; vertex \( (2, 9) \), a maximum. \( x \)-intercepts where \( (x-2)^2 = 9 \), so \( x = 5 \) and \( x = -1 \). \( y \)-intercept at \( x = 0 \): \( -(4) + 9 = 5 \). Max at \( (2,9) \); roots \( -1 \) and 5

  14. A student finds the axis of symmetry of \( y = x^2 - 4x + 1 \) as \( x = -2 \). Find the error.
    Show the full solution

    They dropped the minus in the formula. \( b = -4 \), so \( -\frac{b}{2a} = -\frac{-4}{2} = 2 \). Sketching confirms it: the roots are \( 2 \pm \sqrt{3} \), whose midpoint is 2. \( x = 2 \)

  15. A student says \( y = (x-3)^2 \) has two \( x \)-intercepts. Is that right?
    Show the full solution

    No. Setting it to zero gives \( x = 3 \) twice (a repeated root) so the parabola touches the axis at one point rather than crossing it twice. Its discriminant is zero. One \( x \)-intercept, at \( (3,0) \)

Topic 2.4 · Unit 2 · CA CCSS A-CED.1, F-IF.4

Quadratic modeling and applications

Word problems are where quadratics get tested hardest, because you have to build the equation yourself and then decide which of the two solutions the question actually wants. The algebra is the easy half.

The method
  1. Define the variable in words before writing anything, "let \( w \) be the width in meters". Half the errors in this topic are solving for the wrong quantity.
  2. Translate into an equation and rearrange to standard form. Area problems give a product; projectile problems give a height formula; profit problems give price × quantity.
  3. Match the question to the feature: "when does it hit the ground" → \( x \)-intercept, set \( h = 0 \). "maximum height" or "best price" → the vertex. "when is it at height 20" → set the expression equal to 20.
  4. Solve, then test both answers against the context. Discard negative lengths, negative times before the start, and quantities that exceed a stated limit.
  5. Answer in a sentence with units. A bare number rarely earns full marks on a modeling question.

Where marks are lost: on a "maximum height" question, the vertex gives both numbers and students report the wrong one. The \( x \)-coordinate is when it peaks; the \( y \)-coordinate is how high. Read which the question asked for.

Worked examples

Example 1: maximum of a projectile. A ball's height is \( h = -5t^2 + 30t + 2 \) meters after \( t \) seconds. Find the maximum height and when it occurs.

Maximum means the vertex. \( t = -\frac{30}{2(-5)} = 3 \) seconds. Then \( h = -5(9) + 90 + 2 = 47 \) meters.

So it peaks at 3 seconds, reaching 47 m. Reporting "3 meters" would be the classic error, 3 is a time.

Example 2: an area problem. A rectangular pen uses 40 m of fencing on three sides, with a wall as the fourth. What width gives the maximum area?

Let the width perpendicular to the wall be \( w \). Two widths and one length use the fencing, so the length is \( 40 - 2w \). Area:

\( A = w(40 - 2w) = -2w^2 + 40w \)

Vertex: \( w = -\frac{40}{2(-2)} = 10 \) m, giving \( A = 10 \times 20 = 200 \) m². The maximum area is 200 m² with a width of 10 m.

Example 3: choosing between two solutions. Using \( h = -5t^2 + 30t + 2 \), when is the ball at 42 m?

Set \( -5t^2 + 30t + 2 = 42 \), so \( -5t^2 + 30t - 40 = 0 \). Divide by \( -5 \): \( t^2 - 6t + 8 = 0 \), giving \( (t-2)(t-4) = 0 \) and \( t = 2 \) or \( t = 4 \).

Both are valid: the ball passes 42 m going up at 2 s and again coming down at 4 s. Here you keep both answers, unlike a length problem, where a negative would be discarded.

Practice · 15 problems

1–5 identify what to find, 6–11 solve, 12–15 interpret and diagnose.

  1. To find when a projectile lands, which feature do you need?
    Show the full solution

    Landing means height zero. The \( x \)-intercept

  2. To find a maximum profit, which feature do you need?
    Show the full solution

    The vertex

  3. A rectangle has width \( w \) and length \( w + 5 \). Write its area.
    Show the full solution

    \( A = w^2 + 5w \)

  4. For \( h = -5t^2 + 20t \), find when the object lands.
    Show the full solution

    \( -5t(t-4) = 0 \), so \( t = 0 \) or \( t = 4 \); \( t = 0 \) is the launch. 4 seconds

  5. For \( h = -5t^2 + 20t \), find the time of maximum height.
    Show the full solution

    \( t = -\frac{20}{-10} = 2 \). 2 seconds

  6. For that same model, find the maximum height.
    Show the full solution

    At \( t = 2 \): \( -20 + 40 = 20 \). 20 meters

  7. A rectangle's length is 4 m more than its width and its area is 45 m². Find the width.
    Show the full solution

    \( w(w+4) = 45 \), so \( w^2 + 4w - 45 = 0 \) and \( (w+9)(w-5) = 0 \). A width cannot be \( -9 \). 5 meters

  8. Two consecutive positive integers have a product of 72. Find them.
    Show the full solution

    \( n(n+1) = 72 \), so \( n^2 + n - 72 = 0 \) and \( (n+9)(n-8) = 0 \). Positive means \( n = 8 \). 8 and 9

  9. A farmer has 60 m of fencing for a rectangular pen against a wall, fencing three sides. Write the area as a function of the width \( w \).
    Show the full solution

    Length is \( 60 - 2w \), so \( A = w(60 - 2w) \). \( A = -2w^2 + 60w \)

  10. Using that model, find the width giving maximum area.
    Show the full solution

    \( w = -\frac{60}{2(-2)} = 15 \). 15 meters

  11. And the maximum area?
    Show the full solution

    \( A = 15(60 - 30) = 450 \). 450 m²

  12. For \( h = -5t^2 + 25t \), when is the object at 30 m?
    Show the full solution

    \( -5t^2 + 25t - 30 = 0 \); divide by \( -5 \): \( t^2 - 5t + 6 = 0 \), so \( (t-2)(t-3) = 0 \). Both are valid, rising at 2 s, falling at 3 s. 2 s and 3 s

  13. For that model, what is the greatest height reached, and why can it never be at 35 m?
    Show the full solution

    Vertex at \( t = 2.5 \), giving \( h = -31.25 + 62.5 = 31.25 \) m. Since the maximum is 31.25 m, 35 m is above the entire path, setting \( h = 35 \) would give a negative discriminant. 31.25 m; 35 m is unreachable

  14. A student finds the vertex of a height model at \( (4, 60) \) and answers "the maximum height is 4 meters". Find the error.
    Show the full solution

    They reported the \( x \)-coordinate. The vertex says it peaks at 4 seconds at 60 meters, so the maximum height is 60 m and 4 s is when it occurs. 60 meters, at 4 seconds

  15. A student solves an area problem and reports both \( w = 6 \) and \( w = -11 \). What should they do?
    Show the full solution

    Both satisfy the equation, but a width cannot be negative, so \( -11 \) is rejected on contextual grounds. State the rejection explicitly, examiners look for it. \( w = 6 \); reject \( -11 \)

Unit 2 mixed review · 10 problems · all topics

Unit 2 mixed review: Quadratic Functions and Equations

Part of the work here is choosing the method. Nothing tells you whether to factor, take roots, complete the square or use the formula, decide before you start.

  1. Solve \( x^2 - 7x + 12 = 0 \).
    Show the full solution

    Factors as \( (x-3)(x-4) = 0 \). \( x = 3 \) or \( x = 4 \)

  2. Solve \( 2x^2 = 50 \).
    Show the full solution

    \( x^2 = 25 \), so \( x = \pm 5 \). Dropping the negative root is the standard error. \( x = \pm 5 \)

  3. Solve \( (x - 4)^2 = 9 \).
    Show the full solution

    \( x - 4 = \pm 3 \), so \( x = 7 \) or \( x = 1 \). \( x = 7, 1 \)

  4. Solve \( x^2 + 6x - 3 = 0 \) by completing the square.
    Show the full solution

    \( x^2 + 6x = 3 \); add \( 9 \) to both sides: \( (x+3)^2 = 12 \). \( x = -3 \pm 2\sqrt{3} \). \( x = -3 \pm 2\sqrt{3} \)

  5. Solve \( 3x^2 - 5x - 2 = 0 \) using the formula.
    Show the full solution

    \( x = \frac{5 \pm \sqrt{25 + 24}}{6} = \frac{5 \pm 7}{6} \), giving \( 2 \) and \( -\frac{1}{3} \). \( x = 2, -\frac{1}{3} \)

  6. Find the discriminant of \( x^2 - 4x + 7 = 0 \) and say what it tells you.
    Show the full solution

    \( 16 - 28 = -12 \). Negative, so there are no real solutions and the parabola never crosses the \( x \)-axis. \( -12 \); no real roots

  7. Find the vertex of \( y = x^2 - 6x + 5 \).
    Show the full solution

    \( x = -\frac{b}{2a} = 3 \); then \( y = 9 - 18 + 5 = -4 \). \( (3, -4) \)

  8. State the vertex and axis of symmetry of \( y = -2(x+1)^2 + 8 \).
    Show the full solution

    Vertex form: the sign inside flips, so the vertex is \( (-1, 8) \) and the axis is \( x = -1 \). The negative leading coefficient means it opens downward, so 8 is a maximum. \( (-1, 8) \), \( x = -1 \)

  9. A ball's height is \( h = -16t^2 + 48t + 4 \) feet. Find its maximum height.
    Show the full solution

    \( t = -\frac{48}{2(-16)} = 1.5 \) s. Then \( h = -16(2.25) + 72 + 4 = -36 + 76 = 40 \). 40 feet at 1.5 s

  10. A student solves \( x^2 = 6x \) by dividing both sides by \( x \) to get \( x = 6 \). Find the error.
    Show the full solution

    Dividing by \( x \) assumes \( x \neq 0 \) and discards a root. Move everything to one side and factor: \( x(x-6) = 0 \). \( x = 0 \) or \( x = 6 \)

Topic 3.1 · Unit 3 · CA CCSS G-SRT.2–3

Similarity and similar triangles

Similar figures have the same shape and possibly different size: corresponding angles are congruent and corresponding sides are in a constant ratio. Almost every problem is one proportion, and the marks are in setting it up with matching parts on top.

The method
  1. Similarity criteria for triangles. AA: two pairs of congruent angles is enough, because the third follows from the 180° sum. SSS~: all three side ratios equal. SAS~: two side ratios equal with the included angles congruent.
  2. Congruence is the special case where the ratio is 1. AA works for similarity but AAA never proves congruence, which is the distinction to keep straight.
  3. Write the similarity statement in corresponding order. \( \triangle ABC \sim \triangle DEF \) claims \( A \leftrightarrow D \), \( B \leftrightarrow E \), \( C \leftrightarrow F \), and every proportion you write must respect it.
  4. Set up the proportion consistently: put both sides of one triangle on top, or corresponding sides on top, but do not mix. A reliable habit is \( \frac{\text{small}}{\text{large}} = \frac{\text{small}}{\text{large}} \).
  5. Scale factor \( k \): lengths scale by \( k \), perimeters by \( k \), and areas by \( k^2 \). That last one is tested constantly.

Where marks are lost: a question gives similar triangles with a scale factor of 3 and asks for the ratio of areas. The answer is 9, not 3. Doubling every length of a rectangle quadruples its area, check it on a 2×3 against a 4×6.

Worked examples

Example 1: finding a missing side. \( \triangle ABC \sim \triangle DEF \) with \( AB = 6 \), \( BC = 8 \) and \( DE = 9 \). Find \( EF \).

The statement pairs \( AB \) with \( DE \) and \( BC \) with \( EF \). So

\( \frac{AB}{DE} = \frac{BC}{EF} \), that is \( \frac{6}{9} = \frac{8}{EF} \).

Cross-multiplying: \( 6 \cdot EF = 72 \), so \( EF = 12 \). Check the scale factor: the second triangle is \( \frac{9}{6} = 1.5 \) times the first, and \( 8 \times 1.5 = 12 \) ✓.

Example 2: proving similarity. Two triangles share an angle, and a line parallel to one side cuts the other two. Why are they similar?

The shared angle is congruent to itself. The parallel line makes corresponding angles congruent where the transversal crosses. That is two pairs of congruent angles, so the triangles are similar by AA. This configuration (a line parallel to a side ) is the most common way similarity appears on exams.

Example 3: the area ratio. Two similar triangles have a scale factor of \( \frac{2}{5} \). The smaller has area 12 cm². Find the larger.

Areas scale by the square of the factor, so the ratio of areas is \( \left(\frac{2}{5}\right)^2 = \frac{4}{25} \).

\( \frac{12}{A} = \frac{4}{25} \), so \( 4A = 300 \) and \( A = 75 \) cm². Using \( \frac{2}{5} \) directly would give 30, which is the trap.

Practice · 15 problems

1–6 criteria and ratios, 7–11 missing sides, 12–15 areas and diagnosis.

  1. Two triangles have two pairs of congruent angles. Are they similar?
    Show the full solution

    The third pair follows from the 180° sum. Yes, by AA

  2. Two triangles have three pairs of congruent angles. Are they congruent?
    Show the full solution

    Similar, but the size is unconstrained. Not necessarily congruent

  3. \( \triangle ABC \sim \triangle DEF \). Which side corresponds to \( AC \)?
    Show the full solution

    \( A \to D \) and \( C \to F \). \( DF \)

  4. Two similar figures have a scale factor of 4. What is the ratio of their perimeters?
    Show the full solution

    Perimeter is a length, so it scales by \( k \). 4

  5. Two similar figures have a scale factor of 4. What is the ratio of their areas?
    Show the full solution

    Areas scale by \( k^2 \). 16

  6. A triangle with sides 3, 4, 5 is similar to one with sides 9, 12, 15. Find the scale factor.
    Show the full solution

    \( 9 \div 3 = 3 \), and the other pairs agree. 3

  7. \( \triangle ABC \sim \triangle DEF \), \( AB = 4 \), \( DE = 10 \), \( BC = 6 \). Find \( EF \).
    Show the full solution

    \( \frac{4}{10} = \frac{6}{EF} \), so \( 4 \cdot EF = 60 \) and \( EF = 15 \). 15

  8. \( \triangle PQR \sim \triangle STU \), \( PQ = 15 \), \( ST = 5 \), \( SU = 7 \). Find \( PR \).
    Show the full solution

    The scale factor from \( STU \) to \( PQR \) is \( 15 \div 5 = 3 \), so \( PR = 7 \times 3 = 21 \). 21

  9. A 6 ft person casts a 4 ft shadow while a tree casts a 30 ft shadow. Find the tree's height.
    Show the full solution

    The triangles are similar (same sun angle, both vertical): \( \frac{6}{4} = \frac{h}{30} \), so \( 4h = 180 \) and \( h = 45 \). 45 ft

  10. A photo 4 in by 6 in is enlarged so the short side becomes 10 in. Find the new long side.
    Show the full solution

    Scale factor \( 10 \div 4 = 2.5 \), so \( 6 \times 2.5 = 15 \). 15 in

  11. Two similar triangles have areas 9 cm² and 49 cm². Find the scale factor.
    Show the full solution

    The area ratio is \( k^2 = \frac{49}{9} \), so \( k = \frac{7}{3} \). Take the square root, the ratio of lengths is not \( \frac{49}{9} \). \( \frac{7}{3} \)

  12. Two similar rectangles have a scale factor of \( \frac{1}{3} \). The larger has area annotated as 63 cm². Find the smaller.
    Show the full solution

    Area ratio \( \left(\frac{1}{3}\right)^2 = \frac{1}{9} \), so \( 63 \div 9 = 7 \). 7 cm²

  13. Why does AA prove similarity but AAA fail to prove congruence?
    Show the full solution

    Angles fix the shape but say nothing about size. Two triangles can have identical angles at any scale, so angle information alone can never pin down the lengths that congruence requires. Angles fix shape, not size

  14. A student is told two similar triangles have a scale factor of 3 and answers that the area ratio is 3. Find the error.
    Show the full solution

    Area involves two dimensions, so it scales by \( k^2 = 9 \). Check on a 1×1 square against a 3×3: areas 1 and 9, not 1 and 3. 9

  15. A student writes \( \frac{AB}{EF} = \frac{BC}{DE} \) for \( \triangle ABC \sim \triangle DEF \). Find the error.
    Show the full solution

    The parts do not correspond. The statement pairs \( AB \) with \( DE \) and \( BC \) with \( EF \), so the proportion is \( \frac{AB}{DE} = \frac{BC}{EF} \). Mixing the correspondence gives a wrong answer even with correct arithmetic. \( \frac{AB}{DE} = \frac{BC}{EF} \)

Topic 3.2 · Unit 3 · CA CCSS G-SRT.4–5

Proportional segments in triangles

Three results follow from similarity and appear constantly: a line parallel to a side splits the other two proportionally, an angle bisector splits the opposite side proportionally, and the altitude to the hypotenuse of a right triangle creates geometric means.

The method
  1. Side splitter theorem. If a line parallel to one side of a triangle crosses the other two, it divides them proportionally: \( \frac{\text{top left}}{\text{bottom left}} = \frac{\text{top right}}{\text{bottom right}} \). The converse also holds, equal ratios prove the line is parallel.
  2. Be careful which segments you use. The theorem compares the pieces of each side, not a piece against the whole. Mixing those is the main error here, so label the diagram before writing anything.
  3. Angle bisector theorem. A bisector from a vertex divides the opposite side in the ratio of the two adjacent sides.
  4. Geometric mean, right triangles. The altitude to the hypotenuse creates two smaller triangles similar to each other and to the original. The altitude is the geometric mean of the two hypotenuse pieces: \( h = \sqrt{pq} \). Each leg is the geometric mean of the whole hypotenuse and the piece adjacent to it.
  5. Parallel lines cut by transversals divide them proportionally too, the same idea extended beyond a single triangle.

Where marks are lost: in the side splitter, using \( \frac{\text{piece}}{\text{whole}} \) on one side and \( \frac{\text{piece}}{\text{piece}} \) on the other. Either convention works if applied to both sides consistently; mixing them does not.

Worked examples

Example 1: side splitter. In \( \triangle ABC \), \( \overline{DE} \) is parallel to \( \overline{BC} \) with \( D \) on \( \overline{AB} \) and \( E \) on \( \overline{AC} \). Given \( AD = 6 \), \( DB = 9 \) and \( AE = 8 \), find \( EC \).

Pieces against pieces on both sides: \( \frac{AD}{DB} = \frac{AE}{EC} \), that is \( \frac{6}{9} = \frac{8}{EC} \).

Cross-multiplying: \( 6 \cdot EC = 72 \), so \( EC = 12 \).

Example 2: angle bisector. In \( \triangle ABC \), the bisector from \( A \) meets \( \overline{BC} \) at \( D \). Given \( AB = 8 \), \( AC = 12 \) and \( BD = 6 \), find \( DC \).

The bisector splits \( BC \) in the ratio of the adjacent sides: \( \frac{AB}{AC} = \frac{BD}{DC} \), so \( \frac{8}{12} = \frac{6}{DC} \).

\( 8 \cdot DC = 72 \), giving \( DC = 9 \). Note the larger piece sits against the longer side, which is a useful sanity check.

Example 3: geometric mean. In a right triangle, the altitude to the hypotenuse divides it into pieces of 4 and 9. Find the altitude.

\( h = \sqrt{4 \times 9} = \sqrt{36} = 6 \).

The legs follow the same pattern against the whole hypotenuse of 13: one leg is \( \sqrt{13 \times 4} = 2\sqrt{13} \) and the other \( \sqrt{13 \times 9} = 3\sqrt{13} \). Checking with Pythagoras: \( 52 + 117 = 169 = 13^2 \) ✓.

Practice · 15 problems

1–6 side splitter, 7–10 angle bisector, 11–15 geometric mean and diagnosis.

  1. \( \overline{DE} \parallel \overline{BC} \), \( AD = 4 \), \( DB = 6 \), \( AE = 6 \). Find \( EC \).
    Show the full solution

    \( \frac{4}{6} = \frac{6}{EC} \), so \( 4 \cdot EC = 36 \). \( EC = 9 \)

  2. \( \overline{DE} \parallel \overline{BC} \), \( AD = 5 \), \( DB = 10 \), \( EC = 8 \). Find \( AE \).
    Show the full solution

    \( \frac{5}{10} = \frac{AE}{8} \), so \( AE = 4 \). 4

  3. \( \overline{DE} \parallel \overline{BC} \), \( AD = 3 \), \( DB = 9 \), \( AE = 2 \). Find \( EC \).
    Show the full solution

    \( \frac{3}{9} = \frac{2}{EC} \), so \( EC = 6 \). 6

  4. In a triangle, \( AD = 4 \), \( DB = 6 \), \( AE = 6 \), \( EC = 9 \). Is \( \overline{DE} \parallel \overline{BC} \)?
    Show the full solution

    \( \frac{4}{6} = \frac{2}{3} \) and \( \frac{6}{9} = \frac{2}{3} \). Equal ratios, so by the converse the line is parallel. Yes

  5. In a triangle, \( AD = 3 \), \( DB = 5 \), \( AE = 4 \), \( EC = 7 \). Is \( \overline{DE} \parallel \overline{BC} \)?
    Show the full solution

    \( \frac{3}{5} = 0.6 \) but \( \frac{4}{7} \approx 0.571 \). Not equal. No

  6. \( \overline{DE} \parallel \overline{BC} \), \( AD = 6 \), \( AB = 15 \), \( AE = 8 \). Find \( AC \).
    Show the full solution

    Here whole-to-whole is easier: \( \frac{AD}{AB} = \frac{AE}{AC} \), so \( \frac{6}{15} = \frac{8}{AC} \) and \( AC = 20 \). Using wholes on both sides is consistent, which is what matters. 20

  7. An angle bisector from \( A \) meets \( \overline{BC} \) at \( D \). \( AB = 6 \), \( AC = 9 \), \( BD = 4 \). Find \( DC \).
    Show the full solution

    \( \frac{6}{9} = \frac{4}{DC} \), so \( DC = 6 \). 6

  8. Same setup with \( AB = 10 \), \( AC = 15 \), \( DC = 12 \). Find \( BD \).
    Show the full solution

    \( \frac{10}{15} = \frac{BD}{12} \), so \( BD = 8 \). 8

  9. An angle bisector meets the opposite side, splitting it into 5 and 7. The side adjacent to the 5-piece is 15. Find the other adjacent side.
    Show the full solution

    \( \frac{15}{x} = \frac{5}{7} \), so \( 5x = 105 \) and \( x = 21 \). 21

  10. In a triangle with \( AB = 8 \) and \( AC = 8 \), where does the bisector from \( A \) meet \( \overline{BC} \)?
    Show the full solution

    The ratio is \( 1:1 \), so it meets at the midpoint, which is why the bisector of the vertex angle of an isosceles triangle is also a median. At the midpoint

  11. An altitude to the hypotenuse divides it into 3 and 12. Find the altitude.
    Show the full solution

    \( \sqrt{3 \times 12} = \sqrt{36} = 6 \). 6

  12. An altitude to the hypotenuse divides it into 5 and 20. Find the altitude.
    Show the full solution

    \( \sqrt{100} = 10 \). 10

  13. Hypotenuse pieces are 4 and 5. Find the leg adjacent to the 4-piece.
    Show the full solution

    A leg is the geometric mean of the whole hypotenuse and its adjacent piece: \( \sqrt{9 \times 4} = 6 \). 6

  14. A student uses \( \frac{AD}{AB} = \frac{AE}{EC} \) for the side splitter. Find the error.
    Show the full solution

    The left side is piece-over-whole and the right is piece-over-piece. Pick one and apply it to both: either \( \frac{AD}{DB} = \frac{AE}{EC} \) or \( \frac{AD}{AB} = \frac{AE}{AC} \). Inconsistent ratios

  15. A student finds the altitude to a hypotenuse split into 2 and 8 as \( \frac{2 + 8}{2} = 5 \). Find the error.
    Show the full solution

    They used the arithmetic mean instead of the geometric mean. It is \( \sqrt{2 \times 8} = 4 \), not 5. The geometric mean is always the smaller of the two for distinct values. 4

Topic 3.3 · Unit 3 · CA CCSS G-SRT.8

The Pythagorean theorem and special right triangles

Two special triangles come up so often that recognizing them saves the whole calculation. The Pythagorean theorem handles everything else, provided you identify the hypotenuse correctly, which is where most errors start.

The method
  1. Pythagoras: \( a^2 + b^2 = c^2 \), where \( c \) is always the hypotenuse: the side opposite the right angle, and always the longest. It is never one of the \( a \), \( b \) legs.
  2. Finding a leg: rearrange to \( a^2 = c^2 - b^2 \). Subtract, do not add, a common slip when the missing side is a leg rather than the hypotenuse.
  3. The converse classifies triangles. If \( a^2 + b^2 = c^2 \) it is right; if \( > c^2 \), acute; if \( < c^2 \), obtuse. Always test with \( c \) as the longest side.
  4. 45-45-90: the legs are equal and the hypotenuse is \( \text{leg} \times \sqrt{2} \). This is half a square cut along its diagonal.
  5. 30-60-90: the short leg (opposite 30°) is \( x \), the hypotenuse is \( 2x \), and the long leg (opposite 60°) is \( x\sqrt{3} \). Find the short leg first, then build the other two from it.

Where marks are lost: in a 30-60-90, working from the wrong side. If you are given the hypotenuse, halve it to get the short leg before doing anything else. And common triples (3-4-5, 5-12-13, 8-15-17) and their multiples are worth recognizing, since they turn a calculation into a glance.

Worked examples

Example 1: finding a leg. A right triangle has hypotenuse 13 and one leg 5. Find the other leg.

\( a^2 = 13^2 - 5^2 = 169 - 25 = 144 \), so \( a = 12 \). This is the 5-12-13 triple; spotting it skips the arithmetic entirely.

Example 2: 30-60-90 from the hypotenuse. A 30-60-90 triangle has hypotenuse 14. Find both legs.

The hypotenuse is \( 2x \), so \( x = 7 \), that is the short leg, opposite the 30° angle.

The long leg is \( x\sqrt{3} = 7\sqrt{3} \). Going straight from the hypotenuse to the long leg without finding \( x \) first is where this goes wrong.

Example 3: classifying with the converse. Is a triangle with sides 6, 8 and 11 right, acute or obtuse?

The longest is 11, so compare \( 6^2 + 8^2 = 100 \) against \( 11^2 = 121 \). Since \( 100 < 121 \), the triangle is obtuse. Had the sides been 6, 8, 10 the sums would match and it would be right.

Practice · 15 problems

1–6 Pythagoras, 7–11 special triangles, 12–15 classify and diagnose.

  1. Legs 3 and 4. Find the hypotenuse.
    Show the full solution

    \( \sqrt{9+16} = 5 \). 5

  2. Legs 9 and 12. Find the hypotenuse.
    Show the full solution

    \( \sqrt{81+144} = \sqrt{225} = 15 \), a 3-4-5 triple scaled by 3. 15

  3. Hypotenuse 25, one leg 7. Find the other leg.
    Show the full solution

    \( \sqrt{625 - 49} = \sqrt{576} = 24 \). Subtract, since the missing side is a leg. 24

  4. Hypotenuse 17, one leg 8. Find the other leg.
    Show the full solution

    \( \sqrt{289 - 64} = \sqrt{225} = 15 \). 15

  5. Legs 5 and 5. Find the hypotenuse in exact form.
    Show the full solution

    \( \sqrt{50} = 5\sqrt{2} \), a 45-45-90. \( 5\sqrt{2} \)

  6. A ladder 10 m long leans with its base 6 m from a wall. How high does it reach?
    Show the full solution

    \( \sqrt{100 - 36} = 8 \). 8 m

  7. A 45-45-90 triangle has legs of 7. Find the hypotenuse.
    Show the full solution

    \( 7\sqrt{2} \). \( 7\sqrt{2} \)

  8. A 45-45-90 triangle has hypotenuse \( 6\sqrt{2} \). Find a leg.
    Show the full solution

    Divide by \( \sqrt{2} \). 6

  9. A 30-60-90 triangle has short leg 5. Find the hypotenuse and long leg.
    Show the full solution

    Hypotenuse \( 2 \times 5 = 10 \); long leg \( 5\sqrt{3} \). 10 and \( 5\sqrt{3} \)

  10. A 30-60-90 triangle has hypotenuse 20. Find both legs.
    Show the full solution

    Short leg is half the hypotenuse, 10; long leg is \( 10\sqrt{3} \). 10 and \( 10\sqrt{3} \)

  11. A 30-60-90 triangle has long leg \( 9\sqrt{3} \). Find the short leg.
    Show the full solution

    The long leg is \( x\sqrt{3} \), so \( x = 9 \). 9

  12. Classify a triangle with sides 7, 24, 25.
    Show the full solution

    \( 49 + 576 = 625 = 25^2 \). Right

  13. Classify a triangle with sides 5, 6, 7.
    Show the full solution

    \( 25 + 36 = 61 > 49 \). Acute

  14. A student has hypotenuse 10 and leg 6, and computes \( \sqrt{100 + 36} \). Find the error.
    Show the full solution

    They added when the missing side is a leg. It should be \( \sqrt{100 - 36} = 8 \). A quick check: the answer must be smaller than the hypotenuse, and their \( \sqrt{136} \approx 11.7 \) is larger. 8

  15. A student has a 30-60-90 with hypotenuse 12 and says the long leg is \( 12\sqrt{3} \). Find the error.
    Show the full solution

    The \( x\sqrt{3} \) rule uses the short leg, not the hypotenuse. Halve the hypotenuse first: \( x = 6 \), so the long leg is \( 6\sqrt{3} \approx 10.4 \), which is sensibly less than the hypotenuse, their answer exceeded it. \( 6\sqrt{3} \)

Topic 3.4 · Unit 3 · CA CCSS G-SRT.6–8

Right triangle trigonometry

Trigonometry relates an angle to a ratio of sides. Everything in this topic is one decision, which ratio fits what you have and what you want, followed by one calculator step.

The method
  1. Label relative to the angle you are using. The hypotenuse is fixed, but "opposite" and "adjacent" swap depending on which acute angle you work from. Relabel for each angle rather than trusting the first labeling.
  2. SOH-CAH-TOA: \( \sin\theta = \frac{\text{opp}}{\text{hyp}} \), \( \cos\theta = \frac{\text{adj}}{\text{hyp}} \), \( \tan\theta = \frac{\text{opp}}{\text{adj}} \).
  3. To find a side: pick the ratio containing the known side and the unknown one, then solve. If the unknown is in the denominator, cross-multiply rather than guessing.
  4. To find an angle: use the inverse (\( \sin^{-1} \), \( \cos^{-1} \), \( \tan^{-1} \)) on the ratio of the two sides you know.
  5. Angles of elevation and depression are both measured from the horizontal. They are equal for the same line of sight viewed from either end, which lets you move the angle to whichever triangle is easier.

Where marks are lost: the calculator in radians. A sine of 30 should give 0.5; if it gives \( -0.988 \) the mode is wrong. Check this once at the start of any trig work and it will never cost you a paper.

Worked examples

Example 1: finding a side. A right triangle has a 35° angle and a hypotenuse of 20. Find the side opposite the 35°.

Opposite and hypotenuse means sine: \( \sin 35° = \frac{x}{20} \), so \( x = 20 \sin 35° \approx 20 \times 0.5736 = 11.5 \).

Sanity check: the opposite side must be shorter than the hypotenuse, and 11.5 < 20 ✓.

Example 2: the unknown in the denominator. A right triangle has a 40° angle with the adjacent side 15. Find the hypotenuse.

Adjacent and hypotenuse means cosine: \( \cos 40° = \frac{15}{h} \).

The unknown is underneath, so cross-multiply: \( h \cos 40° = 15 \), giving \( h = \frac{15}{\cos 40°} \approx \frac{15}{0.766} = 19.6 \). The hypotenuse is longest, and 19.6 > 15 ✓.

Example 3: finding an angle. A ramp rises 3 m over a horizontal distance of 8 m. Find its angle of elevation.

Opposite and adjacent means tangent: \( \tan\theta = \frac{3}{8} = 0.375 \).

\( \theta = \tan^{-1}(0.375) \approx 20.6° \). The inverse is what converts a ratio back into an angle.

Practice · 15 problems

1–5 identify the ratio, 6–11 find sides, 12–15 find angles and diagnose. Round to one decimal place.

  1. You know the opposite side and the hypotenuse. Which ratio do you use?
    Show the full solution

    Sine

  2. You know the adjacent side and the hypotenuse. Which ratio?
    Show the full solution

    Cosine

  3. You know the opposite and adjacent sides. Which ratio?
    Show the full solution

    Tangent

  4. In a right triangle, the side opposite is 6 and the hypotenuse is 10. Find \( \sin\theta \).
    Show the full solution

    \( \frac{6}{10} = 0.6 \). 0.6

  5. In that same triangle, find \( \tan\theta \).
    Show the full solution

    The adjacent side is \( \sqrt{100-36} = 8 \), so \( \frac{6}{8} = 0.75 \). 0.75

  6. A 30° angle has hypotenuse 12. Find the opposite side.
    Show the full solution

    \( 12 \sin 30° = 12 \times 0.5 = 6 \). Matches the 30-60-90 rule. 6

  7. A 50° angle has hypotenuse 9. Find the adjacent side.
    Show the full solution

    \( 9 \cos 50° \approx 9 \times 0.643 = 5.8 \). 5.8

  8. A 25° angle has adjacent side 14. Find the opposite side.
    Show the full solution

    \( 14 \tan 25° \approx 14 \times 0.466 = 6.5 \). 6.5

  9. A 60° angle has opposite side 10. Find the hypotenuse.
    Show the full solution

    \( \sin 60° = \frac{10}{h} \), so \( h = \frac{10}{0.866} \approx 11.5 \). 11.5

  10. A 42° angle has adjacent side 20. Find the hypotenuse.
    Show the full solution

    \( h = \frac{20}{\cos 42°} \approx \frac{20}{0.743} = 26.9 \). 26.9

  11. A ladder reaches 12 m up a wall at 70° to the ground. How long is the ladder?
    Show the full solution

    The wall is opposite the 70°: \( \frac{12}{\sin 70°} \approx \frac{12}{0.940} = 12.8 \). 12.8 m

  12. A right triangle has legs 5 and 12. Find the angle opposite the 5.
    Show the full solution

    \( \tan^{-1}\!\left(\frac{5}{12}\right) \approx 22.6° \). 22.6°

  13. A right triangle has hypotenuse 15 and one leg 9. Find the angle opposite that leg.
    Show the full solution

    \( \sin^{-1}\!\left(\frac{9}{15}\right) = \sin^{-1}(0.6) \approx 36.9° \). 36.9°

  14. A student computes \( \sin 30° \) and gets \( -0.988 \). What is wrong?
    Show the full solution

    The calculator is in radian mode, that is \( \sin \) of 30 radians. In degrees \( \sin 30° = 0.5 \) exactly, which is the quickest way to test the mode. Switch to degrees

  15. A student finds a side as \( 20 \times \cos 40° = 15.3 \) when the 20 was the adjacent side and the hypotenuse was wanted. Find the error.
    Show the full solution

    They multiplied when the unknown is in the denominator. \( \cos 40° = \frac{20}{h} \) gives \( h = \frac{20}{\cos 40°} \approx 26.1 \). Their answer was smaller than the adjacent side, which is impossible for a hypotenuse. 26.1

Unit 3 mixed review · 10 problems · all topics

Unit 3 mixed review: Similarity and Right Triangles

Watch which triangle each ratio belongs to, and check that your answer is plausible before you write it, a leg longer than the hypotenuse is always wrong.

  1. Two triangles have angles \( 40^\circ, 60^\circ \) and \( 60^\circ, 80^\circ \). Are they similar?
    Show the full solution

    The first has a third angle of \( 80^\circ \), the second \( 40^\circ \). Both are \( 40 \), \( 60 \), \( 80 \). Yes, by AA

  2. Similar triangles have a scale factor of \( \frac{3}{5} \). The smaller has a side of 12. Find the matching side of the larger.
    Show the full solution

    \( \frac{3}{5} = \frac{12}{x} \), so \( 3x = 60 \). 20

  3. Two similar figures have a scale factor of 2. How do their areas compare?
    Show the full solution

    Area scales by the square of the factor. 4 times

  4. In a triangle, \( DE \parallel BC \) with \( AD = 4 \), \( DB = 6 \), \( AE = 6 \). Find \( EC \).
    Show the full solution

    \( \frac{4}{6} = \frac{6}{EC} \), so \( 4 \cdot EC = 36 \). \( EC = 9 \)

  5. Find the hypotenuse of a right triangle with legs 9 and 12.
    Show the full solution

    \( 81 + 144 = 225 \). 15

  6. A right triangle has hypotenuse 26 and one leg 10. Find the other leg.
    Show the full solution

    \( 676 - 100 = 576 \), so the leg is 24. Adding instead of subtracting is the error here, the hypotenuse is given, not missing. 24

  7. In a \( 30^\circ\text{-}60^\circ\text{-}90^\circ \) triangle the short leg is 7. Find the other two sides.
    Show the full solution

    Long leg \( = 7\sqrt{3} \), hypotenuse \( = 14 \). \( 7\sqrt{3} \) and 14

  8. A \( 45^\circ\text{-}45^\circ\text{-}90^\circ \) triangle has hypotenuse \( 10\sqrt{2} \). Find a leg.
    Show the full solution

    Divide by \( \sqrt{2} \). 10

  9. A ladder leans at \( 68^\circ \) to the ground and reaches 15 feet up a wall. Find its length.
    Show the full solution

    The wall is opposite, the ladder is the hypotenuse, so use sine: \( \sin 68^\circ = \frac{15}{L} \), \( L = \frac{15}{0.9272} \approx 16.2 \). about 16.2 feet

  10. A student finds an angle using \( \sin \theta = \frac{5}{3} \). What went wrong?
    Show the full solution

    Sine cannot exceed 1 in a right triangle, because the opposite side cannot be longer than the hypotenuse. The ratio was inverted; it should be \( \frac{3}{5} \), giving \( \theta \approx 36.9^\circ \). Ratio inverted

Topic 4.1 · Unit 4 · CA CCSS G-C.1–2

Circle vocabulary, central angles and arcs

Circle problems are mostly vocabulary followed by one short calculation. Getting the names right is not pedantry here, the theorems in the next topic are stated in these terms, and you cannot apply a theorem you cannot identify.

The method
  1. The parts. A radius joins the center to the circle. A chord joins two points on the circle. A diameter is a chord through the center, and is twice the radius. A secant is a line through two points of the circle; a tangent touches at exactly one point.
  2. Central angle: vertex at the center. Its arc has the same measure as the angle; this is the definition of arc measure, not a theorem to prove.
  3. Arc measure against arc length. Measure is in degrees and depends only on the angle. Length is a distance and depends on the radius too. A 90° arc is 90° in any circle but far longer in a big one.
  4. Minor, major, semicircle. A minor arc is under 180° and is named with two letters; a major arc is over 180° and needs three letters to distinguish it. A semicircle is exactly 180°.
  5. Arcs around a circle sum to 360°, which is how most missing-arc questions are solved.

Where marks are lost: naming a major arc with two letters. \( \overarc{AB} \) means the minor arc; the major arc needs a point in between, such as \( \overarc{ACB} \). Without it the answer is ambiguous and scores nothing.

Worked examples

Example 1: missing arc. A circle is divided into three arcs measuring \( 4x \), \( 5x \) and \( 3x \). Find each.

Arcs around a circle total 360°: \( 12x = 360 \), so \( x = 30 \).

The arcs are 120°, 150° and 90°. Check: they sum to 360 ✓.

Example 2: central angle and its major arc. A central angle measures 110°. Find the minor arc and the major arc.

The minor arc equals the central angle, so 110°. The major arc is the rest of the circle: \( 360 - 110 = 250° \). Both are needed when a question says "find both arcs".

Example 3: diameter and chord. A circle has radius 13 and a chord of length 24. How far is the chord from the center?

Drop a perpendicular from the center to the chord; it bisects the chord, giving a leg of 12. The radius to an endpoint is the hypotenuse, 13.

\( d = \sqrt{169 - 144} = \sqrt{25} = 5 \). The perpendicular-bisects-the-chord fact is what turns a circle problem into a right-triangle problem, and it is worth reaching for first.

Practice · 15 problems

1–6 vocabulary, 7–11 arcs and angles, 12–15 chords and diagnosis.

  1. What is a chord?
    Show the full solution

    A segment joining two points on the circle

  2. How does a tangent differ from a secant?
    Show the full solution

    A tangent meets the circle once; a secant crosses it twice. One point of contact against two

  3. A circle has radius 7. Find its diameter.
    Show the full solution

    14

  4. A circle has diameter 26. Find its radius.
    Show the full solution

    13

  5. What is the measure of a semicircle?
    Show the full solution

    180°

  6. Why does a major arc need three letters?
    Show the full solution

    Two points define two arcs. Two letters is taken to mean the minor one, so a third point on the arc is needed to specify the major. To distinguish it from the minor arc

  7. A central angle measures 75°. Find its intercepted arc.
    Show the full solution

    Arc measure equals the central angle. 75°

  8. A minor arc measures 140°. Find the major arc.
    Show the full solution

    \( 360 - 140 \). 220°

  9. Three arcs measure \( 2x \), \( 3x \) and \( 5x \). Find \( x \).
    Show the full solution

    \( 10x = 360 \). \( x = 36 \)

  10. Two arcs of a circle measure 85° and 130°. Find the third.
    Show the full solution

    \( 360 - 85 - 130 = 145 \). 145°

  11. Do a 60° arc in a circle of radius 2 and a 60° arc in a circle of radius 10 have the same measure? The same length?
    Show the full solution

    Same measure, both 60°. Different lengths, since length depends on the radius. Same measure, different length

  12. A radius perpendicular to a chord meets it. What does it do to the chord?
    Show the full solution

    It bisects it, which creates two congruent right triangles. Bisects the chord

  13. A circle has radius 10 and a chord 4 units from the center. Find the chord's length.
    Show the full solution

    Half-chord \( = \sqrt{100 - 16} = \sqrt{84} = 2\sqrt{21} \), so the chord is \( 4\sqrt{21} \approx 18.3 \). \( 4\sqrt{21} \)

  14. A circle has radius 5 and a chord of length 8. Find the distance from the center.
    Show the full solution

    Half the chord is 4, so \( \sqrt{25 - 16} = 3 \). 3

  15. A student says a 90° arc is always 90 units long. Find the error.
    Show the full solution

    They confused measure with length. The measure is 90° in every circle, but the length is \( \frac{90}{360} \) of the circumference, which depends on the radius. Measure is not length

Topic 4.2 · Unit 4 · CA CCSS G-C.2–4

Inscribed angles, chords and tangents

One theorem does most of the work in this topic, and one consequence of it appears on almost every exam. The rest is a short list of facts about tangents.

The method
  1. Inscribed angle theorem. An angle with its vertex on the circle is half its intercepted arc. Compare with a central angle, which equals its arc, the halving is the whole difference and the main source of errors.
  2. Same arc, same angle. Two inscribed angles intercepting the same arc are congruent, wherever their vertices sit on the circle.
  3. Angle in a semicircle is 90°. An inscribed angle on a diameter intercepts a 180° arc, and half of 180 is 90. This turns any circle problem with a diameter into a right-triangle problem.
  4. Cyclic quadrilateral: if all four vertices lie on the circle, opposite angles are supplementary, each pair intercepts arcs totaling 360°, and half of that is 180°.
  5. Tangents. A tangent is perpendicular to the radius at the point of contact. Two tangents drawn from the same external point are congruent.

Where marks are lost: halving when you should not, or not halving when you should. Ask where the vertex is. On the circle → inscribed → halve. At the center → central → do not halve.

Worked examples

Example 1: inscribed against central. An arc measures 80°. Find the inscribed angle intercepting it, and the central angle intercepting it.

Inscribed: half the arc, \( 40° \). Central: equal to the arc, \( 80° \). Same arc, two different answers, decided entirely by where the vertex sits.

Example 2: the semicircle. \( \overline{AB} \) is a diameter and \( C \) is on the circle. Given \( AC = 6 \) and \( BC = 8 \), find the radius.

\( \angle ACB \) is inscribed on a diameter, so it is a right angle. Then \( AB \) is the hypotenuse: \( \sqrt{36 + 64} = 10 \).

\( AB \) is the diameter, so the radius is 5. Spotting the right angle is what makes this a two-line problem.

Example 3: tangents from a point. Two tangents are drawn from an external point \( P \), touching at \( A \) and \( B \). Given \( PA = 3x + 4 \) and \( PB = 5x - 6 \), find \( PA \).

Tangents from a common external point are congruent, so \( 3x + 4 = 5x - 6 \), giving \( 10 = 2x \) and \( x = 5 \).

\( PA = 3(5) + 4 = 19 \). Check \( PB = 25 - 6 = 19 \) ✓.

Practice · 15 problems

1–6 inscribed angles, 7–11 semicircles and cyclic quadrilaterals, 12–15 tangents and diagnosis.

  1. An inscribed angle intercepts a 100° arc. Find the angle.
    Show the full solution

    Half the arc. 50°

  2. An inscribed angle measures 35°. Find its intercepted arc.
    Show the full solution

    Double it. 70°

  3. A central angle intercepts a 64° arc. Find the angle.
    Show the full solution

    Central angles equal their arc, no halving. 64°

  4. Two inscribed angles intercept the same arc, and one is 42°. Find the other.
    Show the full solution

    They are congruent. 42°

  5. An inscribed angle intercepts a semicircle. Find the angle.
    Show the full solution

    Half of 180°. 90°

  6. An inscribed angle measures \( 3x \) and its arc is \( 8x - 20 \). Find \( x \).
    Show the full solution

    The arc is double the angle: \( 8x - 20 = 6x \), so \( 2x = 20 \) and \( x = 10 \). \( x = 10 \)

  7. \( \overline{AB} \) is a diameter and \( C \) is on the circle, with \( AC = 5 \) and \( BC = 12 \). Find \( AB \).
    Show the full solution

    \( \angle ACB = 90° \), so \( \sqrt{25 + 144} = 13 \). 13

  8. In that circle, find the radius.
    Show the full solution

    Half the diameter. 6.5

  9. A cyclic quadrilateral has one angle of 95°. Find the angle opposite it.
    Show the full solution

    Opposite angles are supplementary. 85°

  10. A cyclic quadrilateral has angles \( x \) and \( 3x \) opposite each other. Find \( x \).
    Show the full solution

    \( 4x = 180 \). \( x = 45 \)

  11. Why is an angle inscribed in a semicircle always right?
    Show the full solution

    Its intercepted arc is the semicircle, 180°, and an inscribed angle is half its arc. Half of 180° is 90°

  12. A tangent touches a circle at \( T \). What is the angle between the tangent and the radius \( OT \)?
    Show the full solution

    90°

  13. Two tangents from \( P \) touch at \( A \) and \( B \), with \( PA = 14 \). Find \( PB \).
    Show the full solution

    Tangents from a common external point are congruent. 14

  14. A tangent from \( P \) touches at \( T \), with \( PT = 8 \) and radius 6. Find the distance from \( P \) to the center.
    Show the full solution

    The radius is perpendicular to the tangent, giving a right triangle with legs 8 and 6: \( \sqrt{64 + 36} = 10 \). 10

  15. A student sees an inscribed angle intercepting a 120° arc and answers 120°. Find the error.
    Show the full solution

    They applied the central angle rule. The vertex is on the circle, so the angle is half the arc: 60°. Ask where the vertex is before choosing the rule. 60°

Topic 4.3 · Unit 4 · CA CCSS G-C.5, G-GPE.1

Arc length, sector area and equations of circles

Arc length and sector area are both fractions of the whole circle, taken in the same way. The equation of a circle then puts circles onto the coordinate plane, where completing the square comes back from your algebra work.

The method
  1. The fraction is always \( \frac{\theta}{360} \). Arc length is that fraction of the circumference; sector area is that fraction of the area. Same fraction, two different wholes.
  2. Arc length \( = \frac{\theta}{360} \times 2\pi r \). Sector area \( = \frac{\theta}{360} \times \pi r^2 \).
  3. Keep \( \pi \) exact unless a decimal is requested. An answer of \( 6\pi \) is usually preferred to 18.85.
  4. Equation of a circle: \( (x - h)^2 + (y - k)^2 = r^2 \) has center \( (h, k) \) and radius \( r \). The signs flip, exactly as with every other horizontal and vertical shift, and the right side is \( r^2 \), not \( r \).
  5. From general form, complete the square in \( x \) and in \( y \) separately to recover the center and radius.

Where marks are lost: reading the radius straight off the equation. In \( (x-2)^2 + (y+3)^2 = 25 \) the radius is 5, not 25, and the center is \( (2, -3) \), not \( (-2, 3) \). Both slips are easy and both are avoidable by writing the form out first.

Worked examples

Example 1: arc length and sector area together. A circle has radius 9. Find the length of a 60° arc and the area of the 60° sector.

The fraction is \( \frac{60}{360} = \frac{1}{6} \).

Circumference \( = 18\pi \), so the arc is \( \frac{1}{6} \times 18\pi = 3\pi \).

Area \( = 81\pi \), so the sector is \( \frac{1}{6} \times 81\pi = 13.5\pi \).

Example 2: reading a circle equation. Give the center and radius of \( (x + 4)^2 + (y - 1)^2 = 36 \).

Match the form: \( x + 4 = x - (-4) \), so \( h = -4 \); \( k = 1 \). The right side is \( r^2 = 36 \), so \( r = 6 \).

Center \( (-4, 1) \), radius 6.

Example 3: completing the square. Find the center and radius of \( x^2 + y^2 - 6x + 4y - 12 = 0 \).

Group and move the constant: \( (x^2 - 6x) + (y^2 + 4y) = 12 \).

Half of \( -6 \) is \( -3 \), squared 9. Half of 4 is 2, squared 4. Add both to both sides:

\( (x - 3)^2 + (y + 2)^2 = 12 + 9 + 4 = 25 \).

Center \( (3, -2) \), radius 5. Adding to the right side as well is the step students forget.

Practice · 15 problems

1–6 arcs and sectors, 7–11 circle equations, 12–15 completing the square and diagnosis. Leave \( \pi \) in answers.

  1. A circle has radius 6. Find its circumference.
    Show the full solution

    \( 2\pi(6) \). \( 12\pi \)

  2. A circle has radius 6. Find its area.
    Show the full solution

    \( \pi(36) \). \( 36\pi \)

  3. Find the length of a 90° arc in a circle of radius 8.
    Show the full solution

    \( \frac{1}{4} \times 16\pi = 4\pi \). \( 4\pi \)

  4. Find the area of a 90° sector in a circle of radius 8.
    Show the full solution

    \( \frac{1}{4} \times 64\pi = 16\pi \). \( 16\pi \)

  5. Find the length of a 120° arc in a circle of radius 15.
    Show the full solution

    \( \frac{1}{3} \times 30\pi = 10\pi \). \( 10\pi \)

  6. Find the area of a 45° sector in a circle of radius 4.
    Show the full solution

    \( \frac{45}{360} = \frac{1}{8} \), so \( \frac{1}{8} \times 16\pi = 2\pi \). \( 2\pi \)

  7. Give the center and radius of \( (x - 3)^2 + (y - 5)^2 = 49 \).
    Show the full solution

    Center \( (3, 5) \), radius 7

  8. Give the center and radius of \( (x + 2)^2 + (y - 7)^2 = 16 \).
    Show the full solution

    \( x + 2 = x - (-2) \). Center \( (-2, 7) \), radius 4

  9. Write the equation of a circle with center \( (0, 0) \) and radius 10.
    Show the full solution

    \( x^2 + y^2 = 100 \)

  10. Write the equation of a circle with center \( (-1, 4) \) and radius 3.
    Show the full solution

    \( (x + 1)^2 + (y - 4)^2 = 9 \)

  11. Give the center and radius of \( (x - 6)^2 + y^2 = 20 \).
    Show the full solution

    \( r = \sqrt{20} = 2\sqrt{5} \), and \( y^2 \) means \( k = 0 \). Center \( (6, 0) \), radius \( 2\sqrt{5} \)

  12. Find the center and radius of \( x^2 + y^2 - 8x + 2y + 8 = 0 \).
    Show the full solution

    \( (x^2 - 8x) + (y^2 + 2y) = -8 \). Add 16 and 1 to both sides: \( (x-4)^2 + (y+1)^2 = 9 \). Center \( (4, -1) \), radius 3

  13. Find the center and radius of \( x^2 + y^2 + 10x - 4y + 13 = 0 \).
    Show the full solution

    \( (x^2 + 10x) + (y^2 - 4y) = -13 \). Add 25 and 4: \( (x+5)^2 + (y-2)^2 = 16 \). Center \( (-5, 2) \), radius 4

  14. A student says \( (x-2)^2 + (y+3)^2 = 25 \) has center \( (-2, 3) \) and radius 25. Find both errors.
    Show the full solution

    The signs flip, so the center is \( (2, -3) \). And the right side is \( r^2 \), so the radius is \( \sqrt{25} = 5 \). Center \( (2, -3) \), radius 5

  15. A student completing the square on \( x^2 - 6x + y^2 = 7 \) writes \( (x-3)^2 + y^2 = 7 \). Find the error.
    Show the full solution

    They added 9 on the left to form the perfect square but not on the right. It should be \( (x-3)^2 + y^2 = 16 \), giving radius 4. Right side becomes 16

Unit 4 mixed review · 10 problems · all topics

Unit 4 mixed review: Circles

Almost every circle problem reduces to identifying which angle type you are looking at. Name it first, then apply the rule.

  1. A central angle measures \( 84^\circ \). Find its intercepted arc.
    Show the full solution

    A central angle equals its arc. \( 84^\circ \)

  2. An inscribed angle intercepts an arc of \( 110^\circ \). Find the angle.
    Show the full solution

    Half the arc. \( 55^\circ \)

  3. An inscribed angle measures \( 38^\circ \). Find its intercepted arc.
    Show the full solution

    Double it. \( 76^\circ \)

  4. An angle is inscribed in a semicircle. Find its measure.
    Show the full solution

    Half of \( 180^\circ \). \( 90^\circ \)

  5. A tangent meets a radius at the point of tangency. Find the angle between them.
    Show the full solution

    \( 90^\circ \)

  6. Two chords intersect inside a circle; the pieces of one are 4 and 9, and one piece of the other is 6. Find the remaining piece.
    Show the full solution

    \( 4 \times 9 = 6x \), so \( x = 6 \). 6

  7. Find the arc length of a \( 90^\circ \) arc in a circle of radius 8.
    Show the full solution

    \( \frac{90}{360} \times 2\pi(8) = \frac{1}{4}(16\pi) = 4\pi \approx 12.57 \). \( 4\pi \)

  8. Find the area of a \( 60^\circ \) sector in a circle of radius 6.
    Show the full solution

    \( \frac{60}{360} \times \pi(36) = 6\pi \approx 18.85 \). \( 6\pi \)

  9. Write the equation of the circle with center \( (-2, 5) \) and radius 4.
    Show the full solution

    The signs inside flip, and the radius is squared: \( (x+2)^2 + (y-5)^2 = 16 \). \( (x+2)^2 + (y-5)^2 = 16 \)

  10. A student reads \( (x-3)^2 + (y+1)^2 = 25 \) as center \( (-3, 1) \), radius 25. Correct both errors.
    Show the full solution

    The center is read with the signs reversed from what appears inside, so it is \( (3, -1) \); and the right side is \( r^2 \), so \( r = 5 \). Center \( (3,-1) \), radius 5

Topic 5.1 · Unit 5 · CA CCSS G-GMD.1, G-MG.1

Area of polygons and composite figures

The formulas are short and the difficulty is elsewhere: using the correct height, and breaking an awkward shape into pieces you already know. Both are decisions made before any arithmetic.

The method
  1. The formulas. Rectangle \( bh \). Triangle \( \frac{1}{2}bh \). Parallelogram \( bh \). Trapezoid \( \frac{1}{2}(b_1 + b_2)h \). Circle \( \pi r^2 \). Regular polygon \( \frac{1}{2}ap \), with \( a \) the apothem and \( p \) the perimeter.
  2. Height means perpendicular height, always. In a slanted parallelogram or an obtuse triangle the height is not the slanted side, and it may fall outside the figure. Using a slant length is the single most common error in this topic.
  3. Composite figures: decompose into rectangles, triangles and circle pieces, find each area, then add. If a piece is missing (a hole, a notch) find the whole and subtract.
  4. Check units. Area is always square units. Convert lengths to a common unit before multiplying, not after.
  5. Work backwards when given the area: substitute into the formula and solve for the missing length.

Where marks are lost: the trapezoid's \( \frac{1}{2} \). The formula averages the two parallel sides and multiplies by the height; forgetting the half doubles the answer. A quick sanity check is that a trapezoid's area must sit between the two rectangles built on its shorter and longer bases.

Worked examples

Example 1: the wrong height. A parallelogram has a base of 10 cm, a slanted side of 8 cm, and a perpendicular height of 6 cm. Find its area.

\( A = bh = 10 \times 6 = 60 \) cm². The 8 cm is the slanted side and plays no part; it is there precisely to see whether you use the perpendicular height. Answering 80 uses the wrong length.

Example 2: a composite figure. A rectangle 12 m by 8 m has a semicircle of diameter 8 m attached to one short end. Find the total area.

Rectangle: \( 12 \times 8 = 96 \) m².

Semicircle: diameter 8 means radius 4, so \( \frac{1}{2}\pi(4^2) = 8\pi \approx 25.1 \) m².

Total \( \approx 121.1 \) m². Halving the diameter before using it is the step to watch.

Example 3: working backwards. A triangle has area 54 cm² and base 12 cm. Find its height.

\( 54 = \frac{1}{2}(12)h = 6h \), so \( h = 9 \) cm. Substituting into the formula and solving is more reliable than trying to rearrange it from memory.

Practice · 15 problems

1–6 direct formulas, 7–11 composites and reverse problems, 12–15 harder cases and diagnosis.

  1. Find the area of a rectangle 7 cm by 9 cm.
    Show the full solution

    63 cm²

  2. Find the area of a triangle with base 10 and height 6.
    Show the full solution

    \( \frac{1}{2}(10)(6) \). 30

  3. Find the area of a parallelogram with base 15 and perpendicular height 4.
    Show the full solution

    60

  4. Find the area of a trapezoid with parallel sides 6 and 10, height 5.
    Show the full solution

    \( \frac{1}{2}(6+10)(5) = \frac{1}{2}(16)(5) = 40 \). 40

  5. Find the area of a circle of radius 5, leaving \( \pi \).
    Show the full solution

    \( 25\pi \)

  6. Find the area of a circle of diameter 12, leaving \( \pi \).
    Show the full solution

    Radius 6 first. \( 36\pi \)

  7. A triangle has area 40 and height 8. Find its base.
    Show the full solution

    \( 40 = \frac{1}{2}b(8) = 4b \), so \( b = 10 \). 10

  8. A rectangle has area 84 cm² and width 6 cm. Find its length.
    Show the full solution

    14 cm

  9. A regular hexagon has side 6 and apothem \( 3\sqrt{3} \). Find its area.
    Show the full solution

    Perimeter \( = 36 \), so \( A = \frac{1}{2}(3\sqrt{3})(36) = 54\sqrt{3} \approx 93.5 \). \( 54\sqrt{3} \)

  10. A 10 by 6 rectangle has a 3 by 2 rectangular hole. Find the remaining area.
    Show the full solution

    \( 60 - 6 = 54 \). 54

  11. An L-shape is a 10 by 8 rectangle with a 4 by 3 corner removed. Find its area.
    Show the full solution

    \( 80 - 12 = 68 \). 68

  12. A square of side 10 has a circle of radius 5 inscribed in it. Find the area outside the circle but inside the square.
    Show the full solution

    \( 100 - 25\pi \approx 100 - 78.5 = 21.5 \). \( 100 - 25\pi \approx 21.5 \)

  13. A trapezoid has area 45, parallel sides 4 and 11. Find its height.
    Show the full solution

    \( 45 = \frac{1}{2}(15)h = 7.5h \), so \( h = 6 \). 6

  14. A student finds the area of a parallelogram with base 12, slanted side 9 and perpendicular height 7 as \( 12 \times 9 = 108 \). Find the error.
    Show the full solution

    They used the slanted side. Area needs the perpendicular height: \( 12 \times 7 = 84 \). The slant is always longer than the height, so using it always overstates the area. 84

  15. A student computes a trapezoid with sides 5 and 9 and height 4 as \( (5+9)(4) = 56 \). Find the error.
    Show the full solution

    They omitted the \( \frac{1}{2} \), which averages the parallel sides. The area is 28. Check: it must lie between \( 5 \times 4 = 20 \) and \( 9 \times 4 = 36 \), and 56 does not. 28

Topic 5.2 · Unit 5 · CA CCSS G-GMD.1–3

Surface area and volume of solids

Nearly every volume formula is either "base area times height" or one third of that. Sorting the solids into those two families reduces a long formula sheet to two ideas plus the sphere.

The method
  1. Prisms and cylinders, \( V = Bh \), where \( B \) is the area of the base. A cylinder is just a prism with a circular base, so \( V = \pi r^2 h \).
  2. Pyramids and cones, \( V = \frac{1}{3}Bh \). Same base, same height, one third the volume. A cone is \( \frac{1}{3}\pi r^2 h \).
  3. Sphere: \( V = \frac{4}{3}\pi r^3 \) and surface area \( 4\pi r^2 \).
  4. Surface area is the total of the faces. For a prism, that is two bases plus the lateral faces. For a cylinder, two circles plus a rectangle whose width is the circumference: \( SA = 2\pi r^2 + 2\pi rh \).
  5. Slant height is not vertical height. A cone's lateral area uses the slant \( \ell \), while its volume uses the perpendicular height \( h \), and \( \ell = \sqrt{r^2 + h^2} \). Mixing them is the main error with cones and pyramids.

Where marks are lost: units. Volume is cubic, surface area square. And scaling: doubling every dimension multiplies surface area by 4 and volume by 8, never by 2.

Worked examples

Example 1: cylinder, both quantities. A cylinder has radius 3 and height 10. Find its volume and surface area.

Volume: \( \pi(9)(10) = 90\pi \approx 282.7 \).

Surface area: two circles \( 2\pi(9) = 18\pi \), plus the curved part \( 2\pi(3)(10) = 60\pi \). Total \( 78\pi \approx 245.0 \).

Example 2: cone with slant height. A cone has radius 6 and vertical height 8. Find its volume and its slant height.

Volume uses the vertical height: \( \frac{1}{3}\pi(36)(8) = 96\pi \approx 301.6 \).

Slant height: \( \ell = \sqrt{36 + 64} = 10 \). Using 10 in the volume formula would overstate it by 25%, the two heights are not interchangeable.

Example 3: working backwards. A sphere has volume \( 36\pi \). Find its radius.

\( \frac{4}{3}\pi r^3 = 36\pi \). Divide both sides by \( \pi \): \( \frac{4}{3}r^3 = 36 \), so \( r^3 = 27 \) and \( r = 3 \).

Canceling \( \pi \) early keeps the arithmetic clean; carrying 3.14159 through makes this much harder than it needs to be.

Practice · 15 problems

1–6 direct, 7–11 surface area and mixed solids, 12–15 reverse problems and diagnosis. Leave \( \pi \) where it appears.

  1. Find the volume of a rectangular prism 4 by 5 by 6.
    Show the full solution

    120

  2. Find the volume of a cylinder with radius 2 and height 9.
    Show the full solution

    \( \pi(4)(9) \). \( 36\pi \)

  3. Find the volume of a cone with radius 3 and height 7.
    Show the full solution

    \( \frac{1}{3}\pi(9)(7) = 21\pi \). \( 21\pi \)

  4. Find the volume of a sphere of radius 3.
    Show the full solution

    \( \frac{4}{3}\pi(27) = 36\pi \). \( 36\pi \)

  5. Find the volume of a pyramid with a 6 by 6 square base and height 10.
    Show the full solution

    \( \frac{1}{3}(36)(10) = 120 \). 120

  6. A cylinder and a cone share a radius and a height. How do their volumes compare?
    Show the full solution

    The cone is one third of the cylinder. Cone is \( \frac{1}{3} \) of the cylinder

  7. Find the surface area of a cube of edge 5.
    Show the full solution

    Six faces of \( 25 \). 150

  8. Find the surface area of a sphere of radius 4.
    Show the full solution

    \( 4\pi(16) \). \( 64\pi \)

  9. Find the surface area of a cylinder with radius 5 and height 10.
    Show the full solution

    \( 2\pi(25) + 2\pi(5)(10) = 50\pi + 100\pi \). \( 150\pi \)

  10. A cone has radius 5 and vertical height 12. Find its slant height.
    Show the full solution

    \( \sqrt{25 + 144} = 13 \). 13

  11. Using that cone, find its volume.
    Show the full solution

    Volume uses the vertical height 12, not the slant: \( \frac{1}{3}\pi(25)(12) = 100\pi \). \( 100\pi \)

  12. A cylinder has volume \( 45\pi \) and radius 3. Find its height.
    Show the full solution

    \( 9\pi h = 45\pi \), so \( h = 5 \). 5

  13. A cube's edges are doubled. What happens to its volume?
    Show the full solution

    Volume scales by \( k^3 = 8 \). Check on edges 1 and 2: volumes 1 and 8. Multiplied by 8

  14. A student computes a cone's volume using its slant height of 13 instead of its vertical height of 12, with radius 5. By how much is the answer wrong?
    Show the full solution

    They get \( \frac{1}{3}\pi(25)(13) = \frac{325\pi}{3} \) instead of \( 100\pi \), too large by \( \frac{25\pi}{3} \). The slant is always the longest of the three, so using it always overstates the volume. Overstated by \( \frac{25\pi}{3} \)

  15. A student gives a volume in cm². What is wrong?
    Show the full solution

    Volume fills three dimensions, so its units are cubic: cm³. Square units describe area. A unit check catches a surprising number of formula errors. Should be cm³

Topic 5.3 · Unit 5 · CA CCSS G-GMD.4, G-MG.2–3

Cross sections, scaling and density modeling

This topic asks you to use geometry on real objects: what shape appears when a solid is sliced, how measurements change under scaling, and how volume converts into mass or cost.

The method
  1. Cross sections. Picture the slice, not the solid. A vertical cut through a cylinder's axis gives a rectangle; a horizontal cut gives a circle. A cone cut horizontally gives a circle, vertically through the apex gives a triangle. A sphere always gives a circle.
  2. Rotating a 2D shape produces a solid: a rectangle spun about a side gives a cylinder, a right triangle spun about a leg gives a cone, a semicircle spun about its diameter gives a sphere.
  3. Scaling, the rule that matters most: multiply every length by \( k \) and areas scale by \( k^2 \), volumes by \( k^3 \).
  4. Density \( = \frac{\text{mass}}{\text{volume}} \), so mass \( = \) density \( \times \) volume. Population density and cost per unit area work the same way, find the volume or area first, then multiply.
  5. Check units all the way through and convert before multiplying. Mixing cm with m inside one calculation is where these problems go wrong.

Where marks are lost: applying a single scale factor to everything. If a model is built at \( \frac{1}{10} \) scale, its surface area is \( \frac{1}{100} \) and its volume \( \frac{1}{1000} \) of the original, which is why a scale model weighs far less than a tenth of the real thing.

Worked examples

Example 1: a cross section. What shape results from slicing a cone vertically through its apex?

The cut passes through the tip and straight down through the base, so the exposed face is bounded by two slant edges and a diameter, a triangle. A horizontal cut would instead give a circle, so the direction of the cut decides the answer entirely.

Example 2: scaling and mass. A statue is scaled up by a factor of 3. If the original weighs 20 kg, what does the enlargement weigh, assuming the same material?

Weight follows volume, which scales by \( k^3 = 27 \).

\( 20 \times 27 = 540 \) kg. Multiplying by 3 to get 60 kg is the standard error and is wrong by a factor of nine.

Example 3: density. A steel cylinder has radius 2 cm and height 10 cm. Steel has density 7.8 g/cm³. Find its mass.

Volume: \( \pi(4)(10) = 40\pi \approx 125.7 \) cm³.

Mass: \( 125.7 \times 7.8 \approx 980 \) g, or about 0.98 kg. Volume first, density second, doing it in the other order has no meaning.

Practice · 15 problems

1–6 cross sections and rotations, 7–11 scaling, 12–15 density and diagnosis.

  1. What shape is a horizontal cross section of a cylinder?
    Show the full solution

    A circle

  2. What shape is a vertical cross section of a cylinder through its axis?
    Show the full solution

    A rectangle

  3. What shape is any cross section of a sphere?
    Show the full solution

    Every slice, at any angle, gives a circle. A circle

  4. What solid results from rotating a right triangle about one of its legs?
    Show the full solution

    A cone

  5. What solid results from rotating a rectangle about one of its sides?
    Show the full solution

    A cylinder

  6. What shape is a horizontal cross section of a square pyramid?
    Show the full solution

    A square, smaller the higher the cut. A square

  7. Lengths are scaled by 5. By what factor does area scale?
    Show the full solution

    \( k^2 \). 25

  8. Lengths are scaled by 5. By what factor does volume scale?
    Show the full solution

    \( k^3 \). 125

  9. A model is built at \( \frac{1}{4} \) scale. What fraction of the original volume is it?
    Show the full solution

    \( \left(\frac{1}{4}\right)^3 = \frac{1}{64} \). \( \frac{1}{64} \)

  10. A cube of edge 2 has its edges tripled. Find the new volume.
    Show the full solution

    Original 8; scaled by \( 27 \), giving 216. Or directly: edge 6, so \( 6^3 = 216 \). 216

  11. Two similar solids have volumes 8 and 125. Find the scale factor of their lengths.
    Show the full solution

    \( k^3 = \frac{125}{8} \), so \( k = \frac{5}{2} \). Take the cube root. \( \frac{5}{2} \)

  12. A block has volume 200 cm³ and density 2.5 g/cm³. Find its mass.
    Show the full solution

    \( 200 \times 2.5 = 500 \). 500 g

  13. A town of 45 000 people covers 15 km². Find its population density.
    Show the full solution

    \( 45000 \div 15 = 3000 \). 3000 people per km²

  14. A statue is scaled up by a factor of 2. A student says its weight doubles. Find the error.
    Show the full solution

    Weight follows volume, which scales by \( k^3 = 8 \). The enlargement weighs eight times as much, not twice. This is why scaled-up structures need disproportionately stronger supports. Eight times

  15. A student finds the mass of a 50 cm³ object with density 3 g/cm³ as \( 50 \div 3 \approx 16.7 \) g. Find the error.
    Show the full solution

    They divided instead of multiplying. Density is mass per unit volume, so mass \( = \) density \( \times \) volume \( = 150 \) g. A unit check settles it: \( \text{cm}^3 \times \frac{\text{g}}{\text{cm}^3} \) leaves grams. 150 g

Unit 5 mixed review · 10 problems · all topics

Unit 5 mixed review: Measurement and Modeling

Check units on every answer, length, square units for area, cubic for volume. A unit mismatch usually means a formula was used in the wrong place.

  1. Find the area of a trapezoid with parallel sides 7 and 11 and height 6.
    Show the full solution

    \( \frac{1}{2}(7+11)(6) = 54 \). 54 square units

  2. Find the area of a regular hexagon with side 6 and apothem \( 3\sqrt{3} \).
    Show the full solution

    \( \frac{1}{2}(36)(3\sqrt{3}) = 54\sqrt{3} \approx 93.5 \). \( 54\sqrt{3} \)

  3. Find the volume of a cylinder with radius 5 and height 12.
    Show the full solution

    \( \pi(25)(12) = 300\pi \approx 942.5 \). \( 300\pi \)

  4. Find the volume of a cone with radius 6 and height 10.
    Show the full solution

    \( \frac{1}{3}\pi(36)(10) = 120\pi \approx 377 \). Forgetting the third is the standard error. \( 120\pi \)

  5. Find the volume of a sphere of radius 3.
    Show the full solution

    \( \frac{4}{3}\pi(27) = 36\pi \approx 113.1 \). \( 36\pi \)

  6. Find the surface area of a cube with edge 5.
    Show the full solution

    \( 6(25) = 150 \). 150 square units

  7. A cylinder is cut by a plane parallel to its base. Name the cross section.
    Show the full solution

    A circle

  8. A rectangle is rotated about one of its sides. Name the solid produced.
    Show the full solution

    A cylinder

  9. A block has volume 250 cm³ and mass 675 g. Find its density.
    Show the full solution

    \( \frac{675}{250} = 2.7 \). 2.7 g/cm³

  10. Two similar cones have a scale factor of 3. A student says the larger has 9 times the volume. Correct them.
    Show the full solution

    Area scales by the square, but volume scales by the cube of the factor, so it is \( 3^3 = 27 \) times. 27 times

Topic 6.1 · Unit 6 · CA CCSS S-CP.1, S-CP.7

Sample spaces, compound events and the addition rule

Probability is counting, carefully. Almost every error comes from either miscounting the sample space or double-counting outcomes that satisfy two conditions at once.

The method
  1. The sample space is every possible outcome. For equally likely outcomes, \( P(A) = \frac{\text{outcomes in } A}{\text{total outcomes}} \). List or tabulate the space when it is small, two dice give 36 outcomes, not 11.
  2. Probabilities run from 0 to 1. An answer outside that range, or a "probability" of 1.4, means an arithmetic error, check before writing it down.
  3. Complement: \( P(\text{not } A) = 1 - P(A) \). When a question says "at least one", the complement is almost always faster than the direct count.
  4. Addition rule: \( P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \). The subtraction removes outcomes counted in both. If the events are mutually exclusive they cannot both happen, that term is zero, and the rule simplifies.
  5. Decide whether events overlap before choosing the form. Drawing a king and drawing a heart overlap, the king of hearts is both. Drawing a king and drawing a queen do not.

Where marks are lost: forgetting the overlap term. For a card that is a king or a heart, \( \frac{4}{52} + \frac{13}{52} = \frac{17}{52} \) is wrong, because the king of hearts was counted twice. It is \( \frac{16}{52} = \frac{4}{13} \).

Worked examples

Example 1: building the sample space. Two fair dice are rolled. Find the probability the total is 7.

The sample space has \( 6 \times 6 = 36 \) equally likely outcomes. Totals of 7 come from \( (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) \), six of them.

\( P = \frac{6}{36} = \frac{1}{6} \). Counting the 11 possible totals instead of the 36 outcomes is the trap: those totals are not equally likely.

Example 2: the addition rule with overlap. One card is drawn from a standard deck. Find the probability it is a face card or a spade.

Face cards: 12. Spades: 13. Both at once (the spade face cards) 3.

\( P = \frac{12}{52} + \frac{13}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26} \).

Example 3: using the complement. A fair coin is tossed four times. Find the probability of at least one head.

Counting "at least one" directly means one, two, three or four heads. The complement is a single case: no heads at all, which is four tails.

\( P(\text{no heads}) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \), so \( P(\text{at least one}) = 1 - \frac{1}{16} = \frac{15}{16} \).

Practice · 15 problems

1–6 basic probability, 7–11 the addition rule, 12–15 complements and diagnosis.

  1. A fair die is rolled. Find \( P(\text{even}) \).
    Show the full solution

    Three of six outcomes. \( \frac{1}{2} \)

  2. A fair die is rolled. Find \( P(\text{greater than 4}) \).
    Show the full solution

    5 and 6. \( \frac{1}{3} \)

  3. A bag has 5 red and 7 blue marbles. Find \( P(\text{red}) \).
    Show the full solution

    \( \frac{5}{12} \). \( \frac{5}{12} \)

  4. One card is drawn from a deck. Find \( P(\text{heart}) \).
    Show the full solution

    \( \frac{13}{52} \). \( \frac{1}{4} \)

  5. How many outcomes are in the sample space for two dice?
    Show the full solution

    \( 6 \times 6 \). 36

  6. Two dice are rolled. Find \( P(\text{total} = 5) \).
    Show the full solution

    \( (1,4),(2,3),(3,2),(4,1) \), four outcomes. \( \frac{1}{9} \)

  7. A card is drawn. Find \( P(\text{king or queen}) \).
    Show the full solution

    Mutually exclusive, so no overlap term: \( \frac{4}{52} + \frac{4}{52} = \frac{8}{52} \). \( \frac{2}{13} \)

  8. A card is drawn. Find \( P(\text{king or heart}) \).
    Show the full solution

    They overlap in the king of hearts: \( \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} \). \( \frac{4}{13} \)

  9. \( P(A) = 0.4 \), \( P(B) = 0.5 \), \( P(A \text{ and } B) = 0.2 \). Find \( P(A \text{ or } B) \).
    Show the full solution

    \( 0.4 + 0.5 - 0.2 = 0.7 \). 0.7

  10. \( P(A) = 0.3 \), \( P(B) = 0.45 \), and the events are mutually exclusive. Find \( P(A \text{ or } B) \).
    Show the full solution

    No overlap term. 0.75

  11. A die is rolled. Find \( P(\text{even or greater than 3}) \).
    Show the full solution

    Even: 2,4,6. Greater than 3: 4,5,6. Overlap: 4,6. \( \frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{4}{6} \). \( \frac{2}{3} \)

  12. \( P(A) = 0.62 \). Find \( P(\text{not } A) \).
    Show the full solution

    \( 1 - 0.62 \). 0.38

  13. Three coins are tossed. Find \( P(\text{at least one tail}) \).
    Show the full solution

    Complement is all heads, \( \frac{1}{8} \), so \( 1 - \frac{1}{8} = \frac{7}{8} \). \( \frac{7}{8} \)

  14. A student computes \( P(\text{face card or spade}) \) as \( \frac{12}{52} + \frac{13}{52} = \frac{25}{52} \). Find the error.
    Show the full solution

    They omitted the overlap. The three spade face cards were counted twice, so subtract \( \frac{3}{52} \), giving \( \frac{22}{52} = \frac{11}{26} \). \( \frac{11}{26} \)

  15. A student reports a probability of 1.3. What does that tell you?
    Show the full solution

    Probabilities lie between 0 and 1, so the answer is impossible, most likely two overlapping events were added without subtracting the overlap. Range-checking every answer catches this instantly. Impossible value

Topic 6.2 · Unit 6 · CA CCSS S-CP.2–6

Conditional probability and independence

A conditional probability restricts attention to a smaller group, which means the denominator changes. That single idea, plus a test for independence, covers the whole topic.

The method
  1. Notation: \( P(A \mid B) \) reads "the probability of \( A \) given \( B \)". The vertical bar means you already know \( B \) happened, so \( B \) becomes the new sample space.
  2. The formula: \( P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)} \). From a two-way table you can skip it and read counts directly: the cell divided by the row or column total.
  3. Order matters. \( P(A \mid B) \) and \( P(B \mid A) \) are different questions with different denominators and usually different answers. Read which group the question restricts to.
  4. Independence test: \( A \) and \( B \) are independent when \( P(A \mid B) = P(A) \), knowing \( B \) changes nothing. Equivalently \( P(A \text{ and } B) = P(A) \cdot P(B) \).
  5. Multiplication rule: \( P(A \text{ and } B) = P(A) \cdot P(B \mid A) \). Only when independent does this simplify to \( P(A) \cdot P(B) \), which matters for drawing with and without replacement.

Where marks are lost: drawing without replacement and multiplying unchanged probabilities. Two aces from a deck is \( \frac{4}{52} \times \frac{3}{51} \), not \( \frac{4}{52} \times \frac{4}{52} \), both the numerator and the denominator drop.

Worked examples

Example 1: from a two-way table. Of 200 students, 120 take Spanish and, of those, 45 also take music. Find the probability a student takes music given they take Spanish.

"Given they take Spanish" restricts to those 120: \( P(\text{music} \mid \text{Spanish}) = \frac{45}{120} = 0.375 \).

Dividing by 200 would answer a different question, the joint probability of both, which is \( \frac{45}{200} = 0.225 \).

Example 2: testing independence. \( P(A) = 0.3 \), \( P(B) = 0.5 \), \( P(A \text{ and } B) = 0.15 \). Are \( A \) and \( B \) independent?

If independent, \( P(A) \cdot P(B) = 0.3 \times 0.5 = 0.15 \), which matches the given value. Yes, independent. Had the joint probability been 0.2, they would not be.

Example 3: without replacement. A bag has 5 red and 3 blue marbles. Two are drawn without replacement. Find the probability both are red.

First draw: \( \frac{5}{8} \). Given a red was removed, 4 red remain of 7: \( \frac{4}{7} \).

\( P = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14} \). The second fraction changed on both top and bottom, which is what "without replacement" means.

Practice · 15 problems

Problems 1–6 use this table: of 300 people, 180 own a car; of the car owners 108 commute by car, and of the 120 non-owners 24 commute by car.

  1. Find \( P(\text{owns a car}) \).
    Show the full solution

    \( \frac{180}{300} = 0.6 \). 0.6

  2. Find \( P(\text{owns a car and commutes by car}) \).
    Show the full solution

    Joint, so divide by the grand total: \( \frac{108}{300} = 0.36 \). 0.36

  3. Find \( P(\text{commutes by car} \mid \text{owns a car}) \).
    Show the full solution

    Restricted to owners: \( \frac{108}{180} = 0.6 \). 0.6

  4. Find \( P(\text{commutes by car} \mid \text{does not own a car}) \).
    Show the full solution

    \( \frac{24}{120} = 0.2 \). 0.2

  5. Find \( P(\text{owns a car} \mid \text{commutes by car}) \).
    Show the full solution

    Total car commuters: \( 108 + 24 = 132 \). So \( \frac{108}{132} \approx 0.818 \). about 0.818

  6. Are car ownership and commuting by car independent here?
    Show the full solution

    \( P(\text{commute} \mid \text{owns}) = 0.6 \) but \( P(\text{commute}) = \frac{132}{300} = 0.44 \). Since \( 0.6 \neq 0.44 \), knowing ownership changes the probability. Not independent

  7. \( P(A) = 0.5 \), \( P(B) = 0.4 \), \( P(A \text{ and } B) = 0.2 \). Find \( P(A \mid B) \).
    Show the full solution

    \( \frac{0.2}{0.4} = 0.5 \). 0.5

  8. Using those values, are \( A \) and \( B \) independent?
    Show the full solution

    \( P(A \mid B) = 0.5 = P(A) \), so knowing \( B \) changes nothing. Independent

  9. \( P(A) = 0.6 \), \( P(B) = 0.3 \), \( P(A \text{ and } B) = 0.24 \). Are they independent?
    Show the full solution

    \( 0.6 \times 0.3 = 0.18 \neq 0.24 \). Not independent

  10. Two cards are drawn with replacement. Find \( P(\text{both hearts}) \).
    Show the full solution

    With replacement the draws are independent: \( \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \). \( \frac{1}{16} \)

  11. Two cards are drawn without replacement. Find \( P(\text{both hearts}) \).
    Show the full solution

    \( \frac{13}{52} \times \frac{12}{51} = \frac{156}{2652} = \frac{1}{17} \). \( \frac{1}{17} \)

  12. A bag has 4 green and 6 yellow marbles. Two are drawn without replacement. Find \( P(\text{both green}) \).
    Show the full solution

    \( \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15} \). \( \frac{2}{15} \)

  13. Explain in one sentence why \( P(A \mid B) \) and \( P(B \mid A) \) usually differ.
    Show the full solution

    They share a numerator but divide by different totals (one restricts to \( B \), the other to \( A \)) so they agree only when those two are equally likely. Different denominators

  14. A student computes \( P(\text{two aces without replacement}) \) as \( \frac{4}{52} \times \frac{4}{52} \). Find the error.
    Show the full solution

    Without replacement the first ace is gone, so the second draw is \( \frac{3}{51} \). The correct value is \( \frac{12}{2652} = \frac{1}{221} \). \( \frac{1}{221} \)

  15. A student is asked for \( P(\text{music} \mid \text{Spanish}) \) with 45 in the cell and 200 in total, and answers \( \frac{45}{200} \). Find the error.
    Show the full solution

    The condition restricts the sample space to Spanish students, so their count (not the grand total) is the denominator. With 120 Spanish students it is \( \frac{45}{120} = 0.375 \). 0.375

Topic 6.3 · Unit 6 · CA CCSS S-CP.9

Permutations, combinations and counting

When a sample space is too big to list, you count it instead. One question decides which formula applies: does the order matter?

The method
  1. Fundamental counting principle: independent choices multiply. Four shirts and three trousers give \( 4 \times 3 = 12 \) outfits.
  2. Order matters → permutation. \( {}_nP_r = \frac{n!}{(n-r)!} \). Use it for finishing positions, rankings, passwords, arrangements in a row, and assigning distinct roles.
  3. Order does not matter → combination. \( {}_nC_r = \frac{n!}{r!(n-r)!} \). Use it for committees, teams, hands of cards, and choosing a subset.
  4. The test: would swapping two chosen items give a different outcome? A first and second place is different when swapped, so permutation. Two people on the same committee is not, so combination.
  5. Combinations are always smaller than the matching permutation, by a factor of \( r! \). If your combination answer exceeds the permutation, they are the wrong way round.

Where marks are lost: using permutations for committees. Choosing 3 people from 10 for a committee is \( {}_{10}C_3 = 120 \), not \( {}_{10}P_3 = 720 \), the latter counts the same three people in six different orders.

Worked examples

Example 1: permutation. Eight runners race. How many ways can gold, silver and bronze be awarded?

The medals are distinct, so order matters: \( {}_8P_3 = \frac{8!}{5!} = 8 \times 7 \times 6 = 336 \).

Example 2: combination. From 12 students, how many ways to choose a committee of 4?

A committee has no roles, so order does not matter:

\( {}_{12}C_4 = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \frac{11880}{24} = 495 \).

Example 3: counting into a probability. Five cards are dealt from a deck. Find the probability all five are hearts.

Hands are unordered, so both counts are combinations. Favorable: choosing 5 hearts from 13, \( {}_{13}C_5 = 1287 \). Total: \( {}_{52}C_5 = 2{,}598{,}960 \).

\( P = \frac{1287}{2598960} \approx 0.000495 \). Using combinations on top and bottom keeps them consistent, which is what makes the ratio valid.

Practice · 15 problems

1–5 counting principle, 6–11 permutations and combinations, 12–15 probability and diagnosis.

  1. A menu has 5 mains and 4 desserts. How many two-course meals?
    Show the full solution

    \( 5 \times 4 \). 20

  2. A PIN uses 4 digits, repeats allowed. How many PINs?
    Show the full solution

    \( 10^4 \). 10 000

  3. Evaluate \( 5! \).
    Show the full solution

    \( 5 \times 4 \times 3 \times 2 \times 1 \). 120

  4. How many ways can 6 books be arranged on a shelf?
    Show the full solution

    \( 6! = 720 \). 720

  5. Does choosing a president and a treasurer from a club use permutations or combinations?
    Show the full solution

    The roles are distinct, so swapping the two people gives a different outcome. Permutations

  6. Evaluate \( {}_7P_2 \).
    Show the full solution

    \( 7 \times 6 = 42 \). 42

  7. Evaluate \( {}_7C_2 \).
    Show the full solution

    \( \frac{42}{2} = 21 \). 21

  8. From 10 people, how many ways to choose a committee of 3?
    Show the full solution

    \( {}_{10}C_3 = \frac{720}{6} = 120 \). 120

  9. From 10 people, how many ways to award first, second and third prize?
    Show the full solution

    \( {}_{10}P_3 = 720 \). Six times the committee count, because each set of three can be ordered \( 3! = 6 \) ways. 720

  10. How many 5-card hands can be dealt from a deck?
    Show the full solution

    \( {}_{52}C_5 \). 2 598 960

  11. A team of 5 is chosen from 9 players. How many teams?
    Show the full solution

    \( {}_9C_5 = 126 \). 126

  12. Three students are chosen at random from 4 girls and 6 boys. Find the probability all three are girls.
    Show the full solution

    \( \frac{{}_4C_3}{{}_{10}C_3} = \frac{4}{120} = \frac{1}{30} \). \( \frac{1}{30} \)

  13. From 5 red and 4 blue marbles, two are chosen. Find the probability both are blue.
    Show the full solution

    \( \frac{{}_4C_2}{{}_9C_2} = \frac{6}{36} = \frac{1}{6} \). \( \frac{1}{6} \)

  14. A student computes the number of 4-person committees from 12 as \( {}_{12}P_4 \). Find the error.
    Show the full solution

    A committee has no ordering, so this counts each committee \( 4! = 24 \) times. Divide by 24: \( {}_{12}C_4 = 495 \), not 11 880. 495

  15. A student's combination answer is larger than the matching permutation. What does that tell you?
    Show the full solution

    It is impossible, a combination is the permutation divided by \( r! \), so it is always smaller for \( r > 1 \). The two formulas have been swapped. They were swapped

Unit 6 mixed review · 10 problems · all topics

Unit 6 mixed review: Probability

Decide three things before computing: is the sample space what you think it is, do the events overlap, and does order matter.

  1. Two dice are rolled. Find \( P(\text{total} = 9) \).
    Show the full solution

    \( (3,6),(4,5),(5,4),(6,3) \), four of 36. \( \frac{1}{9} \)

  2. A card is drawn. Find \( P(\text{queen or diamond}) \).
    Show the full solution

    \( \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} \). \( \frac{4}{13} \)

  3. Five coins are tossed. Find \( P(\text{at least one head}) \).
    Show the full solution

    Complement: \( 1 - \frac{1}{32} \). \( \frac{31}{32} \)

  4. \( P(A) = 0.55 \), \( P(B) = 0.3 \), \( P(A \text{ and } B) = 0.165 \). Are they independent?
    Show the full solution

    \( 0.55 \times 0.3 = 0.165 \), which matches. Independent

  5. Using those values, find \( P(A \mid B) \).
    Show the full solution

    \( \frac{0.165}{0.3} = 0.55 \), equal to \( P(A) \) as independence requires. 0.55

  6. A bag holds 6 red and 4 blue marbles; two are drawn without replacement. Find \( P(\text{both red}) \).
    Show the full solution

    \( \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} \). \( \frac{1}{3} \)

  7. Evaluate \( {}_8C_3 \).
    Show the full solution

    \( \frac{8 \times 7 \times 6}{6} = 56 \). 56

  8. How many ways can a president, secretary and treasurer be chosen from 9 members?
    Show the full solution

    Distinct roles, so \( {}_9P_3 = 9 \times 8 \times 7 = 504 \). 504

  9. Three students are picked from 5 seniors and 7 juniors. Find the probability all three are seniors.
    Show the full solution

    \( \frac{{}_5C_3}{{}_{12}C_3} = \frac{10}{220} = \frac{1}{22} \). \( \frac{1}{22} \)

  10. A student computes \( P(\text{king or face card}) \) as \( \frac{4}{52} + \frac{12}{52} = \frac{16}{52} \). Find the error.
    Show the full solution

    Every king is a face card, so the kings were counted twice, the overlap is all four of them. Subtracting \( \frac{4}{52} \) leaves \( \frac{12}{52} \), which makes sense: "king or face card" is just "face card." \( \frac{3}{13} \)

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